The function `f(x)=| |x|-1|, x in R`, is differerntiable at all `x in R` except at the points.

Options

A. `1,0,-1`
B. 1
C. `1,-1`
D. `-1`

Explanation

To determine where the function \( f(x) = ||x| - 1| \) is differentiable, we need to analyze the function and its behavior at critical points.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) = ||x| - 1| \) involves absolute values. We need to break it down based on the definition of absolute values.

2. **Breaking Down the Function**:
- For \( x \geq 1 \):
\[ f(x) = |x - 1| = x - 1 \]
- For \( 0 \leq x < 1 \):
\[ f(x) = |1 - x| = 1 - x \]
- For \( -1 < x < 0 \):
\[ f(x) = |1 + x| = 1 + x \]
- For \( x \leq -1 \):
\[ f(x) = |-x - 1| = -x - 1 \]

3. **Identifying Critical Points**:
The critical points where the expression inside the absolute values changes are at \( x = -1, 0, 1 \). We will check the differentiability at these points.

4. **Calculating Left-Hand and Right-Hand Derivatives**:
- **At \( x = 1 \)**:
- Left-hand derivative:
\[ f'(1^-) = \frac{d}{dx}(1 - x) = -1 \]
- Right-hand derivative:
\[ f'(1^+) = \frac{d}{dx}(x - 1) = 1 \]
- Since \( f'(1^-) \neq f'(1^+) \), \( f(x) \) is not differentiable at \( x = 1 \).

- **At \( x = 0 \)**:
- Left-hand derivative:
\[ f'(0^-) = \frac{d}{dx}(1 + x) = 1 \]
- Right-hand derivative:
\[ f'(0^+) = \frac{d}{dx}(1 - x) = -1 \]
- Since \( f'(0^-) \neq f'(0^+) \), \( f(x) \) is not differentiable at \( x = 0 \).

- **At \( x = -1 \)**:
- Left-hand derivative:
\[ f'(-1^-) = \frac{d}{dx}(-x - 1) = -1 \]
- Right-hand derivative:
\[ f'(-1^+) = \frac{d}{dx}(1 + x) = 1 \]
- Since \( f'(-1^-) \neq f'(-1^+) \), \( f(x) \) is not differentiable at \( x = -1 \).

5. **Conclusion**:
The function \( f(x) = ||x| - 1| \) is not differentiable at the points \( x = -1, 0, 1 \).

### Final Answer:
The function \( f(x) = ||x| - 1| \) is differentiable at all \( x \in \mathbb{R} \) except at the points \( -1, 0, 1 \).