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: $L(x) = 16x + 29$

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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.