The derivative of f(x)=|x-1|+|x-3| at x=2 is

Options

A. 2
B. `-2`
C. 0
D. does not exist

Answer

To find the derivative of the function \( f(x) = |x - 1| + |x - 3| \) at \( x = 2 \), we will follow these steps:

### Step 1: Identify the function and its critical points
The function is given as:
\[ f(x) = |x - 1| + |x - 3| \]
The critical points occur where the expressions inside the absolute values change signs, which are at \( x = 1 \) and \( x = 3 \).

### Step 2: Determine the intervals for the piecewise definition
We will analyze the function in the intervals determined by the critical points:
1. For \( x < 1 \): Both \( x - 1 \) and \( x - 3 \) are negative.
\[ f(x) = -(x - 1) - (x - 3) = -x + 1 - x + 3 = -2x + 4 \]
2. For \( 1 \leq x < 3 \): Here, \( x - 1 \) is non-negative and \( x - 3 \) is negative.
\[ f(x) = (x - 1) - (x - 3) = x - 1 - x + 3 = 2 \]
3. For \( x \geq 3 \): Both \( x - 1 \) and \( x - 3 \) are non-negative.
\[ f(x) = (x - 1) + (x - 3) = x - 1 + x - 3 = 2x - 4 \]

### Step 3: Find the derivative in the relevant interval
Since we need to find the derivative at \( x = 2 \), we are in the interval \( 1 \leq x < 3 \), where:
\[ f(x) = 2 \]
The derivative of a constant function is:
\[ f'(x) = 0 \]

### Step 4: Conclusion
Thus, the derivative of \( f(x) \) at \( x = 2 \) is:
\[ f'(2) = 0 \]

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

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