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

$L(x) = f(a) + f'(a)(x - a)$

Given: $f(x) = x + \frac{1}{x}, \quad a = -1$

Step 1: Evaluate $f(a)$

We first find $f(-1)$:

$f(-1) = -1 + \frac{1}{-1} = -1 - 1 = -2$

Step 2: Compute the derivative $f'(x)$

We differentiate $f(x) = x + \frac{1}{x}$ with respect to $x$:

$f'(x) = 1 - \frac{1}{x^2}$

Step 3: Evaluate $f'(a)$

Now, substitute $a = -1$ into the derivative:

$f'(-1) = 1 - \frac{1}{(-1)^2} = 1 - 1 = 0$

Step 4: Write the linearization $L(x)$

Using the formula $L(x) = f(a) + f'(a)(x - a)$, we substitute $f(-1) = -2$ and $f'(-1) = 0$:

$L(x) = -2 + 0 \cdot (x + 1) = -2$

Thus, the linearization is:

$L(x) = -2$

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Do you have any questions or need further explanation?

Here are five related questions to deepen your understanding:

1. What does the linearization of a function represent?
2. How is the derivative used in the linearization process?
3. What happens when $f'(a) = 0$ in the linearization formula?
4. How would the linearization change if $a = 1$ instead of $-1$?
5. Why is linearization useful for approximating functions near a given point?

Tip: Linearization gives a good approximation of the function near the point $x = a$, but the further you move from $a$, the less accurate it becomes.