To solve this problem, we need to find the linearization $L(x)$ of the function $f(x)$ at $x = a$.

1. Function and Point of Linearization:

We are given: $f(x) = x + \frac{1}{x}, \quad a = 1.$ We will linearize $f(x)$ around $x = 1$.

2. Formula for Linearization:

The linearization of a function $f(x)$ at $x = a$ is given by the formula: $L(x) = f(a) + f'(a)(x - a).$ So, we need to calculate $f(a)$ and $f'(a)$, where $a = 1$.

3. Step 1: Calculate $f(a)$:

Substitute $a = 1$ into $f(x)$: $f(1) = 1 + \frac{1}{1} = 2.$

4. Step 2: Calculate the Derivative $f'(x)$:

Differentiate $f(x)$: $f'(x) = \frac{d}{dx}\left(x + \frac{1}{x}\right) = 1 - \frac{1}{x^2}.$ Now, evaluate $f'(x)$ at $x = 1$: $f'(1) = 1 - \frac{1}{1^2} = 0.$

5. Step 3: Write the Linearization $L(x)$:

Using the linearization formula: $L(x) = f(1) + f'(1)(x - 1).$ Substitute $f(1) = 2$ and $f'(1) = 0$: $L(x) = 2 + 0(x - 1) = 2.$

Final Answer:

The linearization of $f(x)$ at $x = 1$ is: $L(x) = 2.$

Would you like further details or clarifications on any part of the solution?

Related Questions:

1. How do you interpret the result of a linearization geometrically?
2. What happens to the linearization if $f'(a) \neq 0$?
3. Can you linearize the function $f(x) = x + \frac{1}{x}$ at another point, say $a = 2$?
4. How does the linearization approximate the function near $x = a$?
5. How would the linearization change if the function had higher-order terms?

Tip:

Linearization is essentially finding the equation of the tangent line to the function at a given point. It approximates the function well when you're close to that point!