Find the linearization​ L(x) at xequalsa.
​f(x)equalsnothingxplusStartFraction 1 Over x EndFraction
​,   aequalsnegative 1
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Part 1
​L(x)equals

enter your response here
​(Simplify your​ answer.)

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.

Find the linearization L(x) at x = a. f(x) = x + 1/x, a = -1. L(x) = ? (Simplify your answer.)

Find the linearization of the function f(x) = x + 1/x at x = -1 using the linear approximation formula. This solution covers key calculus concepts, such as finding derivatives and using them to approximate functions near a given point.

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