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.