Number of points where `f(x)=x^2-|x^2-1|+2||x|-1|+2|x|-7` is non-differentiable is a. 0 b. 1 c. 2 d. 3

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To determine the number of points where the function

\[ f(x) = x^2 - |x^2 - 1| + 2||x| - 1| + 2|x| - 7 \]

is non-differentiable, we will analyze the function step by step, focusing on the points where the absolute value expressions change.

### Step 1: Identify critical points
The function contains absolute values, which can change based on the input \( x \). We need to identify the points where the expressions inside the absolute values equal zero.

1. **For \( |x^2 - 1| \)**:
- \( x^2 - 1 = 0 \) gives \( x = 1 \) and \( x = -1 \).

2. **For \( ||x| - 1| \)**:
- \( |x| - 1 = 0 \) gives \( x = 1 \) and \( x = -1 \) (since \( |x| \) is non-negative).

Thus, the critical points to check are \( x = -1, 0, 1 \).

### Step 2: Break down the function based on intervals
We will analyze the function in the intervals determined by the critical points:

1. **For \( x < -1 \)**:
- \( |x^2 - 1| = x^2 - 1 \)
- \( ||x| - 1| = -|x| + 1 = -(-x) + 1 = x + 1 \)
- Therefore, \( f(x) = x^2 - (x^2 - 1) + 2(x + 1) + 2(-x) - 7 \)
- Simplifying this gives \( f(x) = -4x - 8 \).

2. **For \( -1 \leq x < 0 \)**:
- \( |x^2 - 1| = 1 - x^2 \)
- \( ||x| - 1| = -x - 1 \)
- Therefore, \( f(x) = x^2 - (1 - x^2) + 2(-x - 1) + 2(-x) - 7 \)
- Simplifying this gives \( f(x) = -2x - 6 \).

3. **For \( 0 \leq x < 1 \)**:
- \( |x^2 - 1| = 1 - x^2 \)
- \( ||x| - 1| = x - 1 \)
- Therefore, \( f(x) = x^2 - (1 - x^2) + 2(x - 1) + 2x - 7 \)
- Simplifying this gives \( f(x) = 2x - 6 \).

4. **For \( x \geq 1 \)**:
- \( |x^2 - 1| = x^2 - 1 \)
- \( ||x| - 1| = x - 1 \)
- Therefore, \( f(x) = x^2 - (x^2 - 1) + 2(x - 1) + 2x - 7 \)
- Simplifying this gives \( f(x) = 4x - 8 \).

### Step 3: Find the derivatives in each interval
Now we will find the derivative \( f'(x) \) in each interval:

1. **For \( x < -1 \)**:
- \( f'(x) = -4 \).

2. **For \( -1 < x < 0 \)**:
- \( f'(x) = -2 \).

3. **For \( 0 < x < 1 \)**:
- \( f'(x) = 2 \).

4. **For \( x > 1 \)**:
- \( f'(x) = 4 \).

### Step 4: Evaluate the left-hand and right-hand derivatives at critical points
We need to check the left-hand derivative (LHD) and right-hand derivative (RHD) at the critical points \( x = -1, 0, 1 \):

1. **At \( x = -1 \)**:
- LHD = -4 (from the left)
- RHD = -2 (from the right)
- Since LHD ≠ RHD, \( f(x) \) is non-differentiable at \( x = -1 \).

2. **At \( x = 0 \)**:
- LHD = -2 (from the left)
- RHD = 2 (from the right)
- Since LHD ≠ RHD, \( f(x) \) is non-differentiable at \( x = 0 \).

3. **At \( x = 1 \)**:
- LHD = 2 (from the left)
- RHD = 4 (from the right)
- Since LHD ≠ RHD, \( f(x) \) is non-differentiable at \( x = 1 \).

### Conclusion
The function \( f(x) \) is non-differentiable at three points: \( x = -1, 0, 1 \).

Thus, the number of points where \( f(x) \) is non-differentiable is **3**.

### Final Answer
The correct option is **d. 3**.

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