$$  $$We have graph (C) : $$ f\left(x\right) $$$$ = $$$$  $$$$ -\frac12x $$$$ +1+\ln\frac{x}{x-2} $$set on $$  $$$$ D=\left(-\infty,0\right)\cup\left(2,+\infty\right) $$. Prove that $$ \frac{x}{x-2}<1\text{, x}\in\left(-\infty,0\right) $$and $$ \frac{x}{x-2}>1,\text{ x}\in\left(2,+\infty\right) $$.

To prove that $\frac{x}{x-2}<1$ for $x \in (-\infty, 0)$ and $\frac{x}{x-2}>1$ for $x \in (2, +\infty)$, we will analyze the behavior of the function $\frac{x}{x-2}$ in these intervals.

### Case 1: $x \in (-\infty, 0)$

1. **Function Analysis**: Consider the expression $\frac{x}{x-2}$.

   We need to show that $\frac{x}{x-2}<1$. To do this, we start by setting up the inequality:
   \[
   \frac{x}{x-2} < 1
   \]

2. **Simplifying the Inequality**:
   \[
   \frac{x}{x-2} - 1 < 0
   \]
   \[
   \frac{x - (x-2)}{x-2} < 0
   \]
   \[
   \frac{x - x + 2}{x-2} < 0
   \]
   \[
   \frac{2}{x-2} < 0
   \]

3. **Determining the Sign**: The fraction $\frac{2}{x-2}$ will be negative when the denominator $x-2$ is negative.

   \[
   x - 2 < 0
   \]
   \[
   x < 2
   \]

   Since we are given that $x \in (-\infty, 0)$, which is entirely less than 2, the condition $x < 2$ is always satisfied in this interval. Therefore,
   \[
   \frac{2}{x-2} < 0 \quad \text{for} \quad x \in (-\infty, 0)
   \]

   This means:
   \[
   \frac{x}{x-2} < 1 \quad \text{for} \quad x \in (-\infty, 0)
   \]

### Case 2: $x \in (2, +\infty)$

1. **Function Analysis**: Consider the expression $\frac{x}{x-2}$.

   We need to show that $\frac{x}{x-2}>1$. To do this, we start by setting up the inequality:
   \[
   \frac{x}{x-2} > 1
   \]

2. **Simplifying the Inequality**:
   \[
   \frac{x}{x-2} - 1 > 0
   \]
   \[
   \frac{x - (x-2)}{x-2} > 0
   \]
   \[
   \frac{x - x + 2}{x-2} > 0
   \]
   \[
   \frac{2}{x-2} > 0
   \]

3. **Determining the Sign**: The fraction $\frac{2}{x-2}$ will be positive when the denominator $x-2$ is positive.

   \[
   x - 2 > 0
   \]
   \[
   x > 2
   \]

   Since we are given that $x \in (2, +\infty)$, which is entirely greater than 2, the condition $x > 2$ is always satisfied in this interval. Therefore,
   \[
   \frac{2}{x-2} > 0 \quad \text{for} \quad x \in (2, +\infty)
   \]

   This means:
   \[
   \frac{x}{x-2} > 1 \quad \text{for} \quad x \in (2, +\infty)
   \]

### Conclusion
We have proven that:

- $\frac{x}{x-2} < 1$ for $x \in (-\infty, 0)$
- $\frac{x}{x-2} > 1$ for $x \in (2, +\infty)$

Would you like more details or have any questions?

Here are some more questions you might consider asking:

1. How do I find the derivative of a given function?
2. Can you explain the Fundamental Theorem of Calculus?
3. What is the method for solving a definite integral?
4. How do you determine the limit of a function as $x$ approaches a certain value?
5. What are the different methods for solving differential equations?

Tip: When dealing with inequalities, it's helpful to consider the sign of each term and the intervals where these terms change their sign. This approach simplifies finding the solution set.

Learn how to prove inequalities involving the rational function x/(x-2) for x in (-∞, 0) and x in (2, +∞). This detailed solution analyzes the behavior of the function in different intervals and applies basic algebraic techniques to establish the required inequalities.

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