Let `f(x) = |x- x_(1)| + |x-x_(2)|`, where `x_(1) and x_(2)` are distinct real numbers. Then the number of points at which f(x) is minimum

Options

A. 1
B. 2
C. 3
D. more than 3

Answer

To solve the problem, we need to analyze the function \( f(x) = |x - x_1| + |x - x_2| \), where \( x_1 \) and \( x_2 \) are distinct real numbers. We want to find the number of points at which \( f(x) \) is minimized.

### Step 1: Understand the Function
The function \( f(x) \) is the sum of two absolute values. This means it represents the total distance from the point \( x \) to the points \( x_1 \) and \( x_2 \).

### Step 2: Identify Critical Points
The function \( f(x) \) can change its behavior at the points \( x_1 \) and \( x_2 \). Therefore, we will evaluate \( f(x) \) in the intervals defined by these points:
1. \( x < x_1 \)
2. \( x_1 \leq x < x_2 \)
3. \( x \geq x_2 \)

### Step 3: Analyze Each Interval
1. **For \( x < x_1 \)**:
\[ f(x) = (x_1 - x) + (x_2 - x) = (x_1 + x_2 - 2x) \]
Here, \( f(x) \) is a linear function with a negative slope, which means it is decreasing.

2. **For \( x_1 \leq x < x_2 \)**:
\[ f(x) = (x - x_1) + (x_2 - x) = (x_2 - x_1) \]
In this interval, \( f(x) \) is constant.

3. **For \( x \geq x_2 \)**:
\[ f(x) = (x - x_1) + (x - x_2) = (2x - x_1 - x_2) \]
Here, \( f(x) \) is a linear function with a positive slope, which means it is increasing.

### Step 4: Determine Minimum Points
From the analysis:
- \( f(x) \) is decreasing for \( x < x_1 \).
- \( f(x) \) is constant for \( x_1 \leq x < x_2 \).
- \( f(x) \) is increasing for \( x \geq x_2 \).

The minimum value of \( f(x) \) occurs when \( x \) is between \( x_1 \) and \( x_2 \). Since \( f(x) \) is constant in this interval, the minimum value is achieved at every point in the interval \( [x_1, x_2] \).

### Conclusion
The number of points at which \( f(x) \) is minimum is infinite, as it is constant for all \( x \) in the interval \( [x_1, x_2] \).