Verify the truth of Rolle's Theorem for the following functions
`f(x)=|x-1|" in "[1,2]`

Answer

To verify the truth of Rolle's Theorem for the function \( f(x) = |x - 1| \) on the interval \([1, 2]\), we need to check the three conditions of the theorem:

### Step 1: Check Continuity
The first condition of Rolle's Theorem states that the function must be continuous on the closed interval \([a, b]\).

1. **Function Definition**: The function \( f(x) = |x - 1| \) can be rewritten as:
\[ f(x) = \begin{cases} 1 - x & \text{if } x < 1 \\ x - 1 & \text{if } x \geq 1 \end{cases} \]
Since we are only interested in the interval \([1, 2]\), we can use \( f(x) = x - 1 \) for \( x \in [1, 2] \).

2. **Continuity Check**: The function \( f(x) = x - 1 \) is a linear function, which is continuous everywhere. Therefore, it is continuous on the closed interval \([1, 2]\).

### Step 2: Check Differentiability
The second condition states that the function must be differentiable on the open interval \((a, b)\).

1. **Differentiability Check**: The function \( f(x) = x - 1 \) is differentiable for all \( x \) in the open interval \((1, 2)\). The derivative is:
\[ f'(x) = 1 \]
Since \( f'(x) \) exists for all \( x \) in \((1, 2)\), the function is differentiable on this interval.

### Step 3: Check Endpoints
The third condition requires that \( f(a) = f(b) \).

1. **Evaluate Endpoints**:
- Calculate \( f(1) \):
\[ f(1) = |1 - 1| = 0 \]
- Calculate \( f(2) \):
\[ f(2) = |2 - 1| = 1 \]

2. **Check Equality**:
\[ f(1) = 0 \quad \text{and} \quad f(2) = 1 \]
Since \( f(1) \neq f(2) \), the third condition is not satisfied.

### Conclusion
Since the third condition of Rolle's Theorem is not satisfied, we conclude that Rolle's Theorem does not apply to the function \( f(x) = |x - 1| \) on the interval \([1, 2]\).