To find the first-order neighborhood of the function $f(x) = \frac{1}{x} + \ln(x)$ around a point $x_0$, you typically expand the function using a first-order Taylor series expansion around that point. The first-order Taylor expansion is given by:

$f(x) \approx f(x_0) + f'(x_0)(x - x_0)$

Step 1: Find $f(x_0)$

Let's choose $x_0 = 1$ for simplicity (since you didn't specify a point). Plugging $x_0 = 1$ into the function:

$f(1) = \frac{1}{1} + \ln(1) = 1 + 0 = 1$

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

First, let's calculate the derivative of the function $f(x) = \frac{1}{x} + \ln(x)$:

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

Step 3: Evaluate the derivative at $x_0 = 1$

Now, evaluate this derivative at $x_0 = 1$:

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

Step 4: Construct the first-order Taylor expansion

The first-order Taylor expansion around $x_0 = 1$ is then:

$f(x) \approx f(1) + f'(1)(x - 1)$

Substituting the values we found:

$f(x) \approx 1 + 0 \cdot (x - 1) = 1$

So, the first-order approximation (neighborhood) of $f(x) = \frac{1}{x} + \ln(x)$ around $x = 1$ is simply $f(x) \approx 1$.

Would you like more details, or do you have any questions?

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Here are 5 related questions that could further your understanding:

1. How does the second-order Taylor expansion of $f(x)$ differ from the first-order expansion?
2. What is the geometric interpretation of a first-order Taylor expansion?
3. Can the function $f(x) = \frac{1}{x} + \ln(x)$ have points of inflection?
4. How would the first-order approximation change if you expanded around a different point, such as $x_0 = 2$?
5. What is the significance of the point where the derivative equals zero in a Taylor series expansion?

Tip: Taylor series expansions are powerful tools for approximating functions near a given point. The more terms you include, the more accurate the approximation becomes, especially near the expansion point.