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To find the linearization \( L(x) \) of a function \( f(x) \) at \( a \), we use the formula:

\[
L(x) = f(a) + f'(a)(x - a)
\]

### Given function
The function is:
\[
f(x) = x^3 - x^2 + 9
\]
and the point \( a = -2 \).

### Steps to find \( L(x) \):

1. **Calculate \( f(a) \):**

   Substitute \( x = -2 \) into \( f(x) \):
   \[
   f(-2) = (-2)^3 - (-2)^2 + 9 = -8 - 4 + 9 = -3
   \]

   So, \( f(-2) = -3 \).

2. **Find \( f'(x) \):**

   Differentiate \( f(x) \) with respect to \( x \):
   \[
   f'(x) = 3x^2 - 2x
   \]

3. **Calculate \( f'(a) \):**

   Substitute \( x = -2 \) into \( f'(x) \):
   \[
   f'(-2) = 3(-2)^2 - 2(-2) = 3 \cdot 4 + 4 = 12 + 4 = 16
   \]

   So, \( f'(-2) = 16 \).

4. **Write the linearization \( L(x) \):**

   Substitute \( f(a) = -3 \), \( f'(a) = 16 \), and \( a = -2 \) into the linearization formula:
   \[
   L(x) = -3 + 16(x + 2)
   \]

   Simplify the expression:
   \[
   L(x) = -3 + 16x + 32 = 16x + 29
   \]

### Final Answer
The linearization of \( f(x) \) at \( a = -2 \) is:
\[
L(x) = 16x + 29
\]

Would you like more details on any part of this solution?

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### Related Questions
1. How is the derivative used in the process of linearization?
2. What does the linearization represent in terms of the function’s graph?
3. How does changing the point \( a \) affect the linearization?
4. How can linearization be used to approximate values of \( f(x) \) near \( a \)?
5. Can the linearization provide information about the concavity of \( f(x) \)?

### Tip
The linearization is a powerful approximation tool that works best for values of \( x \) close to \( a \). It provides a linear estimate of the function's behavior near that point.

Find the linearization L(x) of the function f(x) = x^3 - x^2 + 9 at a = -2.

Learn how to find the linearization L(x) of the function f(x) = x^3 - x^2 + 9 at the point a = -2. This solution covers key calculus concepts, including differentiation and linear approximation, with a step-by-step approach. Suitable for advanced high school students.

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