Let `f(x) = ||x|-1|,` then points where, `f(x)` is not differentiable is/are

Options

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

Answer

To determine the points where the function \( f(x) = ||x| - 1| \) is not differentiable, we will analyze the function step by step.

### Step 1: Identify the inner function
The function can be broken down as follows:
1. The innermost function is \( |x| \).
2. The next function is \( |x| - 1 \).
3. Finally, we take the absolute value of that result, giving us \( ||x| - 1| \).

### Step 2: Analyze the first absolute value
The function \( |x| \) is defined as:
- \( |x| = x \) for \( x \geq 0 \)
- \( |x| = -x \) for \( x < 0 \)

### Step 3: Analyze the second absolute value
Next, we consider \( |x| - 1 \):
- For \( x \geq 0 \), \( |x| - 1 = x - 1 \)
- For \( x < 0 \), \( |x| - 1 = -x - 1 \)

### Step 4: Determine where \( |x| - 1 = 0 \)
We find the points where \( |x| - 1 = 0 \):
- \( x - 1 = 0 \) gives \( x = 1 \)
- \( -x - 1 = 0 \) gives \( x = -1 \)

Thus, \( |x| - 1 = 0 \) at \( x = -1 \) and \( x = 1 \).

### Step 5: Analyze the outer absolute value
Now we analyze \( ||x| - 1| \):
- For \( x \geq 1 \): \( ||x| - 1| = |x - 1| = x - 1 \)
- For \( 0 \leq x < 1 \): \( ||x| - 1| = |x - 1| = 1 - x \)
- For \( -1 < x < 0 \): \( ||x| - 1| = |-x - 1| = -(-x - 1) = x + 1 \)
- For \( x \leq -1 \): \( ||x| - 1| = |-x - 1| = -(-x - 1) = -x - 1 \)

### Step 6: Identify points of non-differentiability
The function \( f(x) \) is not differentiable at points where there are sharp corners or cusps. We check the points:
1. \( x = -1 \)
2. \( x = 0 \)
3. \( x = 1 \)

### Step 7: Conclusion
The points where \( f(x) \) is not differentiable are:
- \( x = -1 \)
- \( x = 0 \)
- \( x = 1 \)

Thus, the final answer is that \( f(x) \) is not differentiable at the points \( (-1, 0) \), \( (0, 1) \), and \( (1, 0) \).