The value of the derivative of `|x-1| + |x-3|` at x=2 is:

Options

A. `-2`
B. 0
C. 2
D. not defined

Answer

To find the value of the derivative of the function \( f(x) = |x-1| + |x-3| \) at \( x = 2 \), we will analyze the function in different intervals defined by the points where the absolute values change.

### Step-by-Step Solution:

1. **Identify the critical points**:
The absolute value function changes at the points where the expressions inside the absolute values are zero. For \( |x-1| \), this occurs at \( x = 1 \), and for \( |x-3| \), this occurs at \( x = 3 \). Thus, we have critical points at \( x = 1 \) and \( x = 3 \).

2. **Define the function in intervals**:
We will analyze the function in three intervals:
- For \( x < 1 \): Both \( x-1 \) and \( x-3 \) are negative.
\[ f(x) = -(x-1) - (x-3) = -x + 1 - x + 3 = -2x + 4 \]
- For \( 1 \leq x < 3 \): \( x-1 \) is non-negative and \( x-3 \) is negative.
\[ f(x) = (x-1) - (x-3) = x - 1 - x + 3 = 2 \]
- For \( x \geq 3 \): Both \( x-1 \) and \( x-3 \) are non-negative.
\[ f(x) = (x-1) + (x-3) = x - 1 + x - 3 = 2x - 4 \]

3. **Find the derivative in each interval**:
- For \( x < 1 \):
\[ f'(x) = -2 \]
- For \( 1 \leq x < 3 \):
\[ f'(x) = 0 \]
- For \( x \geq 3 \):
\[ f'(x) = 2 \]

4. **Evaluate the derivative at \( x = 2 \)**:
Since \( 2 \) falls in the interval \( 1 \leq x < 3 \), we use the derivative from this interval:
\[ f'(2) = 0 \]

### Final Answer:
The value of the derivative of \( |x-1| + |x-3| \) at \( x = 2 \) is \( 0 \).