The problem asks us to find the maximum value of the function $v(x) = x^3 + 6x^2 + 9x + 2$ from the given multiple-choice options.

Step 1: Identify Critical Points

To find the maximum, compute the derivative of $v(x)$ and solve $v'(x) = 0$: $v'(x) = 3x^2 + 12x + 9$ Set $v'(x) = 0$: $3x^2 + 12x + 9 = 0$ Divide through by 3: $x^2 + 4x + 3 = 0$ Factorize: $(x + 3)(x + 1) = 0$ Thus, the critical points are $x = -3$ and $x = -1$.

Step 2: Determine Maximum or Minimum

Use the second derivative test. Compute $v''(x)$: $v''(x) = 6x + 12$ Evaluate $v''(x)$ at each critical point:

• At $x = -3$: $v''(-3) = 6(-3) + 12 = -18 + 12 = -6 \quad (\text{negative, so } x = -3 \text{ is a local maximum}).$
• At $x = -1$: $v''(-1) = 6(-1) + 12 = -6 + 12 = 6 \quad (\text{positive, so } x = -1 \text{ is a local minimum}).$

Step 3: Find $v(-3)$

Substitute $x = -3$ into $v(x)$: $v(-3) = (-3)^3 + 6(-3)^2 + 9(-3) + 2$ $v(-3) = -27 + 6(9) - 27 + 2$ $v(-3) = -27 + 54 - 27 + 2 = 2$

Thus, the maximum value is $v(-3) = 2$.

Step 4: Verify Answer

The correct choice is: $\boxed{d. \, y_{\text{max}}(-3) = 2}$

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Follow-Up Questions:

1. How do we determine the nature of critical points using the second derivative?
2. What is the significance of the second derivative test in finding maxima or minima?
3. Can we confirm the function's global maximum over an interval? Why or why not?
4. Why does the cubic term influence the overall shape of the function?
5. How can we extend this analysis to find inflection points?

Tip: Always verify critical points using both the first and second derivative tests to ensure accuracy.