Given the function : (1/3)x^3-3x^2-6
Part A: Find the linearization L(x) at x = 3.
Part B: Use this linearization to approximate f(3.01). Is the
approximation greater than or less than the actual value of f(3.01)?
Give a reason for your answer.

### Part A: Finding the Linearization \( L(x) \) at \( x = 3 \)

The linearization \( L(x) \) of a function \( f(x) \) at a point \( x = a \) is given by:
\[
L(x) = f(a) + f'(a)(x - a)
\]

Given \( f(x) = \frac{1}{3}x^3 - 3x^2 - 6 \), let's calculate \( f(3) \) and \( f'(3) \).

1. **Calculate \( f(3) \):**
   \[
   f(3) = \frac{1}{3}(3)^3 - 3(3)^2 - 6 = \frac{1}{3}(27) - 27 - 6 = 9 - 27 - 6 = -24
   \]

2. **Find \( f'(x) \):**
   Differentiate \( f(x) \) with respect to \( x \):
   \[
   f'(x) = \frac{d}{dx} \left(\frac{1}{3}x^3 - 3x^2 - 6\right) = x^2 - 6x
   \]
   Now, calculate \( f'(3) \):
   \[
   f'(3) = (3)^2 - 6(3) = 9 - 18 = -9
   \]

3. **Construct the linearization \( L(x) \):**
   \[
   L(x) = f(3) + f'(3)(x - 3) = -24 - 9(x - 3)
   \]
   Simplifying:
   \[
   L(x) = -9x + 3
   \]

### Part B: Approximate \( f(3.01) \) and Compare with the Actual Value

1. **Using the linearization \( L(x) \) to approximate \( f(3.01) \):**
   \[
   L(3.01) = -9(3.01) + 3 = -27.09 + 3 = -24.09
   \]

2. **Actual value of \( f(3.01) \):**
   Let's calculate the exact value of \( f(3.01) \):
   \[
   f(3.01) = \frac{1}{3}(3.01)^3 - 3(3.01)^2 - 6
   \]
   Expanding the terms:
   \[
   (3.01)^3 = 27.270901
   \]
   \[
   3(3.01)^2 = 27.1803
   \]
   So:
   \[
   f(3.01) = \frac{1}{3} \times 27.270901 - 27.1803 - 6 = 9.0903 - 27.1803 - 6 = -24.09
   \]

### Conclusion: 
The linear approximation \( L(3.01) \) is less than the actual value \( f(3.01) \). Since the function \( f(x) \) is concave down (as \( f''(x) = 2x - 6 < 0 \) for \( x = 3 \)), the linear approximation will always be below the actual value.

Do you need any more details or have any questions? Here are 5 questions that can extend this discussion:

1. How does the concavity of a function affect the accuracy of linear approximations?
2. What is the importance of linearization in calculus and real-world applications?
3. How would the approximation differ if we used a quadratic approximation instead of a linear one?
4. What is the second derivative test, and how could it help determine the concavity at any point?
5. Can you find the linearization of a function at any given point using the same method?

**Tip:** When approximating functions using linearization, always consider the function's concavity to estimate whether your approximation is an under- or overestimate.

Explore the concept of linearization in calculus through a detailed solution. Learn how to find the linearization of a function at a specific point, and use it to approximate f(3.01) for the function f(x) = (1/3)x^3 - 3x^2 - 6. Compare the linear approximation with the actual value, emphasizing the role of concavity in accuracy.

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