If f (x) =`|x-1| +|x-2|` ,then at x=2

Options

A. f (x) is differentiable
B. f(x) is not differentiable
C. `f'(2^(-) ) =2 `
D. ` f'( 2^(+))=0`

Answer

To solve the problem, we need to analyze the function \( f(x) = |x - 1| + |x - 2| \) at \( x = 2 \). We will determine the value of the function and check its differentiability at that point.

### Step 1: Define the function based on intervals
The absolute value function behaves differently based on the value of \( x \). We need to consider the critical points where the expressions inside the absolute values change signs, which are \( x = 1 \) and \( x = 2 \).

1. **For \( x < 1 \)**:
\[ f(x) = -(x - 1) - (x - 2) = -x + 1 - x + 2 = -2x + 3 \]

2. **For \( 1 \leq x < 2 \)**:
\[ f(x) = (x - 1) - (x - 2) = x - 1 - x + 2 = 1 \]

3. **For \( x \geq 2 \)**:
\[ f(x) = (x - 1) + (x - 2) = x - 1 + x - 2 = 2x - 3 \]

### Step 2: Evaluate the function at \( x = 2 \)
Since \( x = 2 \) falls into the third case where \( x \geq 2 \):
\[ f(2) = 2(2) - 3 = 4 - 3 = 1 \]

### Step 3: Find the derivative from both sides
Next, we need to find the derivative of \( f(x) \) to check for differentiability at \( x = 2 \).

1. **For \( x < 2 \)** (from the interval \( 1 \leq x < 2 \)):
\[ f'(x) = 0 \]

2. **For \( x > 2 \)**:
\[ f'(x) = \frac{d}{dx}(2x - 3) = 2 \]

### Step 4: Check for continuity and differentiability
At \( x = 2 \):
- The left-hand derivative \( f'(2^-) = 0 \)
- The right-hand derivative \( f'(2^+) = 2 \)

Since the left-hand derivative does not equal the right-hand derivative, the function is not differentiable at \( x = 2 \).

### Final Conclusion
Thus, we conclude that:
- \( f(2) = 1 \)
- \( f'(2) \) does not exist (the function is not differentiable at \( x = 2 \)).