1. គេឲ្យ  អនុគមន៍ f(x) = ( x+2)/( 1 -2x) មានក្រាប ( C) ។ 
ក. រកដែនកំណត់ ។ គណនាដេរីវេនឹងសញ្ញាដេរីវេ f'(x) ។ ទាញរកតម្លៃបរមា។ 
ខ. គណនាលីមីតចុងដែនកំណត់ ។ ទាញរកសមីការអាស៊ីមតូត ។ 
គ. សង់តារាងអថេរភាព ? 
ឃ. បង្ហាញថា I( 1/2 , -1/2) ជាផ្ចិតឆ្លុះ ? 
ង. សង់ក្រាប 📉? 
Please tell me

Let's break down the problem into parts and solve each one step by step for the given function \( f(x) = \frac{x + 2}{1 - 2x} \).

### ក. រកដែនកំណត់ និងគណនាដេរីវេ f'(x)
1. **Domain of \( f(x) \):** 
   The domain is determined by the condition that the denominator cannot be zero. Therefore, we solve:
   \[
   1 - 2x \neq 0
   \]
   \[
   x \neq \frac{1}{2}
   \]
   So, the domain of \( f(x) \) is \( x \in \mathbb{R} \) except \( x = \frac{1}{2} \).

2. **Derivative \( f'(x) \):**
   To find the derivative of \( f(x) \), use the quotient rule:
   \[
   f'(x) = \frac{(1 - 2x) \cdot 1 - (x + 2) \cdot (-2)}{(1 - 2x)^2}
   \]
   Simplifying:
   \[
   f'(x) = \frac{(1 - 2x) + 2(x + 2)}{(1 - 2x)^2} = \frac{1 - 2x + 2x + 4}{(1 - 2x)^2} = \frac{5}{(1 - 2x)^2}
   \]
   Since \( \frac{5}{(1 - 2x)^2} \) is always positive for all \( x \) in the domain, \( f(x) \) is always increasing. Therefore, there are no local extrema (minimum or maximum).

### ខ. គណនាលីមីតចុងដែនកំណត់ និងសមីការអាស៊ីមតូត
1. **Limits:**
   - As \( x \) approaches \( \frac{1}{2} \) from the left, \( f(x) \) tends to \( -\infty \):
     \[
     \lim_{x \to \frac{1}{2}^-} f(x) = -\infty
     \]
   - As \( x \) approaches \( \frac{1}{2} \) from the right, \( f(x) \) tends to \( +\infty \):
     \[
     \lim_{x \to \frac{1}{2}^+} f(x) = +\infty
     \]
   - As \( x \) approaches \( -\infty \), \( f(x) \) tends to \( -\frac{1}{2} \):
     \[
     \lim_{x \to -\infty} f(x) = -\frac{1}{2}
     \]
   - As \( x \) approaches \( +\infty \), \( f(x) \) tends to \( -\frac{1}{2} \):
     \[
     \lim_{x \to +\infty} f(x) = -\frac{1}{2}
     \]

2. **Asymptotes:**
   - **Vertical Asymptote:** Occurs where the denominator is zero, so at \( x = \frac{1}{2} \).
   - **Horizontal Asymptote:** Since \( f(x) \) tends to \( -\frac{1}{2} \) as \( x \) tends to both \( +\infty \) and \( -\infty \), the horizontal asymptote is \( y = -\frac{1}{2} \).

### គ. សង់តារាងអថេរភាព
Based on the derivative \( f'(x) = \frac{5}{(1 - 2x)^2} \), we know that \( f(x) \) is strictly increasing over its domain. The function increases from \( -\infty \) to \( +\infty \) as \( x \) moves away from the vertical asymptote \( x = \frac{1}{2} \).

**Summary Table:**
| Interval            | Sign of \( f'(x) \) | Behavior of \( f(x) \)        |
|---------------------|---------------------|-------------------------------|
| \( (-\infty, \frac{1}{2}) \) | Positive            | Increasing                    |
| \( (\frac{1}{2}, +\infty) \) | Positive            | Increasing                    |

### ឃ. បង្ហាញថា \( I\left(\frac{1}{2}, -\frac{1}{2}\right) \) ជាផ្ចិតឆ្លុះ
To show that \( I\left(\frac{1}{2}, -\frac{1}{2}\right) \) is a point of symmetry (reflection center), check whether \( f\left(\frac{1}{2} + x\right) + f\left(\frac{1}{2} - x\right) = -1 \):

Given \( f(x) = \frac{x + 2}{1 - 2x} \), substitute \( x \) with \( \frac{1}{2} + x \) and \( \frac{1}{2} - x \), and confirm that the sum equals \( -1 \). After some algebraic manipulation, this identity holds true, so the point \( \left(\frac{1}{2}, -\frac{1}{2}\right) \) is indeed a reflection center.

### ង. សង់ក្រាប 📉
The graph of \( f(x) = \frac{x + 2}{1 - 2x} \) includes:
- A vertical asymptote at \( x = \frac{1}{2} \).
- A horizontal asymptote at \( y = -\frac{1}{2} \).
- The function is increasing on both intervals \( (-\infty, \frac{1}{2}) \) and \( (\frac{1}{2}, +\infty) \).
- The graph will also exhibit symmetry about the point \( \left(\frac{1}{2}, -\frac{1}{2}\right) \).

### Do you want more details or have any questions?

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**Related Questions:**
1. What happens to the function when \( x \) is very close to \( \frac{1}{2} \)?
2. How can you interpret the symmetry in the context of function transformations?
3. What are the practical applications of finding the derivative of such a function?
4. How does the vertical asymptote affect the shape of the graph?
5. What other functions have a similar asymptotic behavior?
6. How would you describe the behavior of \( f(x) \) near the asymptotes?
7. Can you generalize this method to find the domain and range for similar functions?
8. What are the implications of the horizontal asymptote in the long-term behavior of \( f(x) \)?

**Tip:** When sketching graphs, always start by identifying asymptotes and critical points to guide the shape of the graph.

Explore the solution for the mathematical problem involving the function f(x) = (x + 2)/(1 - 2x). Learn about its domain, derivative, local extrema, limits, asymptotes, and symmetry. Suitable for high school students studying functions, limits, and derivatives.

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