Rolle's theorem is applicable for the function `f(x) = |x-1|` in `[0,2]`.

Explanation

To determine whether Rolle's theorem is applicable for the function \( f(x) = |x - 1| \) on the interval \([0, 2]\), we need to check the three conditions of Rolle's theorem:

1. **Continuity on the closed interval \([a, b]\)**.
2. **Differentiability on the open interval \((a, b)\)**.
3. **Equality of function values at the endpoints, i.e., \( f(a) = f(b) \)**.

Let's go through these conditions step by step.

### Step 1: Check Continuity

The function \( f(x) = |x - 1| \) can be expressed as:
- \( f(x) = -(x - 1) = 1 - x \) for \( x < 1 \)
- \( f(x) = x - 1 \) for \( x \geq 1 \)

Since \( |x - 1| \) is a piecewise linear function, it is continuous everywhere, including the interval \([0, 2]\).

**Conclusion**: \( f(x) \) is continuous on \([0, 2]\).

### Step 2: Check Differentiability

Next, we need to check if \( f(x) \) is differentiable on the open interval \((0, 2)\).

- For \( x < 1 \) (i.e., \( (0, 1) \)):
\[ f(x) = 1 - x \implies f'(x) = -1 \]

- For \( x > 1 \) (i.e., \( (1, 2) \)):
\[ f(x) = x - 1 \implies f'(x) = 1 \]

At \( x = 1 \), we need to check the left-hand and right-hand derivatives:
- Left-hand derivative at \( x = 1 \):
\[ f'(1^-) = -1 \]
- Right-hand derivative at \( x = 1 \):
\[ f'(1^+) = 1 \]

Since the left-hand derivative and right-hand derivative at \( x = 1 \) are not equal, \( f(x) \) is not differentiable at \( x = 1 \).

**Conclusion**: \( f(x) \) is not differentiable on \((0, 2)\).

### Step 3: Check Endpoint Values

Now, we check the values of the function at the endpoints:
- \( f(0) = |0 - 1| = 1 \)
- \( f(2) = |2 - 1| = 1 \)

Since \( f(0) = f(2) = 1 \), this condition is satisfied.

### Final Conclusion

Rolle's theorem requires all three conditions to be satisfied. Here, while the function is continuous on \([0, 2]\) and \( f(0) = f(2) \), it is not differentiable at \( x = 1 \). Therefore, we conclude that:

**Rolle's theorem is not applicable for the function \( f(x) = |x - 1| \) on the interval \([0, 2]\).**

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