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We are tasked with finding the relative maximum and minimum values for the function \( f(x) = 2x^3 + x^2 + 15 \) on the interval \([-4, 4]\). Let’s approach this step-by-step.

### 1. **Find the critical points:**
   We first need to find the derivative \( f'(x) \) and solve for when it is equal to zero, as critical points occur where \( f'(x) = 0 \) or where the derivative does not exist.

   \[
   f(x) = 2x^3 + x^2 + 15
   \]

   Taking the derivative:

   \[
   f'(x) = 6x^2 + 2x
   \]

   Set \( f'(x) = 0 \):

   \[
   6x^2 + 2x = 0
   \]

   Factor the equation:

   \[
   2x(3x + 1) = 0
   \]

   Therefore, the critical points are:

   \[
   x = 0 \quad \text{or} \quad x = -\frac{1}{3}
   \]

### 2. **Evaluate the function at the critical points and endpoints:**
   Now we need to evaluate the function \( f(x) \) at the critical points \( x = 0 \) and \( x = -\frac{1}{3} \), as well as at the endpoints \( x = -4 \) and \( x = 4 \).

   \[
   f(0) = 2(0)^3 + (0)^2 + 15 = 15
   \]
   \[
   f\left(-\frac{1}{3}\right) = 2\left(-\frac{1}{3}\right)^3 + \left(-\frac{1}{3}\right)^2 + 15
   \]
   Simplifying this:
   \[
   f\left(-\frac{1}{3}\right) = 2\left(-\frac{1}{27}\right) + \frac{1}{9} + 15 = -\frac{2}{27} + \frac{3}{27} + 15 = \frac{1}{27} + 15 = \frac{406}{27}
   \]

   Finally, calculate at the endpoints:
   \[
   f(4) = 2(4)^3 + (4)^2 + 15 = 2(64) + 16 + 15 = 128 + 16 + 15 = 159
   \]
   \[
   f(-4) = 2(-4)^3 + (-4)^2 + 15 = 2(-64) + 16 + 15 = -128 + 16 + 15 = -97
   \]

### 3. **Compare values:**
   - \( f(0) = 15 \)
   - \( f\left(-\frac{1}{3}\right) = \frac{406}{27} \approx 15.04 \)
   - \( f(4) = 159 \)
   - \( f(-4) = -97 \)

   The maximum occurs at \( x = 4 \) where \( f(4) = 159 \), and the minimum occurs at \( x = -4 \) where \( f(-4) = -97 \).

### 4. **Conclusion:**
   The correct option does not appear in the provided options, as the maximum is at \( f(4) = 159 \) and the minimum is at \( f(-4) = -97 \).

Would you like further clarification or more details on any step?

Here are 5 related questions:
1. How do you find the second derivative to confirm the nature of critical points?
2. What is the geometric significance of critical points on a graph?
3. How do endpoint values influence global extrema on a closed interval?
4. Can there be more than one relative maximum or minimum in cubic functions?
5. How does the behavior of the derivative at a critical point indicate whether it is a max or min?

**Tip:** Always remember to check both the critical points and endpoints when finding extrema on a closed interval.

Find the relative maximum and minimum values for the function f(x) = 2x^3 + x^2 + 15 on the interval [-4, 4].

Discover how to find the relative maximum and minimum for the function f(x) = 2x^3 + x^2 + 15 on the interval [-4, 4] using the derivative. Learn how to evaluate the function at critical points and endpoints, applying calculus concepts like the first derivative test.

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