To find the absolute maximum and minimum values of the given function $f(x) = \frac{1}{2}x^4 - \frac{2}{3}x^3 - 2x^2 + 3$ on the interval $[-3, 3]$, follow these steps:

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Step 1: Take the derivative of $f(x)$

To find critical points, calculate the derivative $f'(x)$: $f'(x) = 2x^3 - 2x^2 - 4x$

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Step 2: Factorize $f'(x)$

Factorize $f'(x)$ to find critical points: $f'(x) = 2x(x^2 - x - 2)$

Factorize $x^2 - x - 2$: $f'(x) = 2x(x - 2)(x + 1)$

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Step 3: Solve $f'(x) = 0$

Set $f'(x) = 0$: $2x(x - 2)(x + 1) = 0$ The critical points are: $x = 0, \, x = 2, \, x = -1$

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Step 4: Evaluate $f(x)$ at critical points and endpoints

Evaluate $f(x)$ at $x = -3, -1, 0, 2, 3$.

• At $x = -3$: $f(-3) = \frac{1}{2}(-3)^4 - \frac{2}{3}(-3)^3 - 2(-3)^2 + 3 = 40.5 + 18 - 18 + 3 = 43.5$

• At $x = -1$: $f(-1) = \frac{1}{2}(-1)^4 - \frac{2}{3}(-1)^3 - 2(-1)^2 + 3 = 0.5 + \frac{2}{3} - 2 + 3 \approx 1.167$

• At $x = 0$: $f(0) = \frac{1}{2}(0)^4 - \frac{2}{3}(0)^3 - 2(0)^2 + 3 = 3$

• At $x = 2$: $f(2) = \frac{1}{2}(2)^4 - \frac{2}{3}(2)^3 - 2(2)^2 + 3 = 8 - \frac{16}{3} - 8 + 3 \approx -2.333$

• At $x = 3$: $f(3) = \frac{1}{2}(3)^4 - \frac{2}{3}(3)^3 - 2(3)^2 + 3 = 40.5 - 18 - 18 + 3 = 7.5$

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Step 5: Identify the absolute maximum and minimum

• Absolute maximum: $f(-3) = 43.5$
• Absolute minimum: $f(2) \approx -2.333$

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Final Answer:

• Maximum: $43.5$ at $x = -3$
• Minimum: $-2.333$ at $x = 2$

Would you like a detailed explanation of any specific step? Here are related questions:

1. How is the derivative of a function calculated?
2. Why do we check endpoints in addition to critical points?
3. How do you factorize a cubic polynomial?
4. What is the difference between local and absolute extrema?
5. How can you confirm extrema using the second derivative test?

Tip: Always evaluate endpoints in a closed interval to ensure all potential extrema are considered.