If `f(x) = |x+1|(|x|+|x-1|)`, then at what point(s) is the function not differentiable over the interval `[-2, 2]`?

Options

A. (a) `-1`
B. (b) `0`
C. (c) `1`
D. (d) `(1)/(2)`

Answer

To determine the points at which the function \( f(x) = |x+1|(|x|+|x-1|) \) is not differentiable over the interval \([-2, 2]\), we will analyze the function by breaking it down into intervals based on the critical points where the absolute values change.

### Step 1: Identify critical points
The critical points for the absolute value functions in \( f(x) \) are:
- From \( |x + 1| \): changes at \( x = -1 \)
- From \( |x| \): changes at \( x = 0 \)
- From \( |x - 1| \): changes at \( x = 1 \)

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

### Step 2: Define the function in different intervals
We will now express \( f(x) \) in different intervals based on the critical points:

1. **Interval \([-2, -1)\)**:
- Here, \( |x + 1| = -(x + 1) = -x - 1 \)
- \( |x| = -x \)
- \( |x - 1| = -x + 1 \)
- Therefore,
\[ f(x) = (-x - 1)(-x + (-x + 1)) = (-x - 1)(-2x + 1) = (x + 1)(2x - 1) = 2x^2 + x - 1 \]

2. **Interval \([-1, 0)\)**:
- Here, \( |x + 1| = x + 1 \)
- \( |x| = -x \)
- \( |x - 1| = -x + 1 \)
- Therefore,
\[ f(x) = (x + 1)(-x + (-x + 1)) = (x + 1)(-2x + 1) = -(x + 1)(2x - 1) = -2x^2 - x + 1 \]

3. **Interval \([0, 1)\)**:
- Here, \( |x + 1| = x + 1 \)
- \( |x| = x \)
- \( |x - 1| = -x + 1 \)
- Therefore,
\[ f(x) = (x + 1)(x + (-x + 1)) = (x + 1)(1) = x + 1 \]

4. **Interval \([1, 2]\)**:
- Here, \( |x + 1| = x + 1 \)
- \( |x| = x \)
- \( |x - 1| = x - 1 \)
- Therefore,
\[ f(x) = (x + 1)(x + (x - 1)) = (x + 1)(2x - 1) = 2x^2 + x - 1 \]

### Step 3: Compute the derivatives in each interval
Now, we will find the derivatives of \( f(x) \) in each of the intervals:

1. **For \( x \in [-2, -1) \)**:
\[ f'(x) = \frac{d}{dx}(2x^2 + x - 1) = 4x + 1 \]

2. **For \( x \in [-1, 0) \)**:
\[ f'(x) = \frac{d}{dx}(-2x^2 - x + 1) = -4x - 1 \]

3. **For \( x \in [0, 1) \)**:
\[ f'(x) = \frac{d}{dx}(x + 1) = 1 \]

4. **For \( x \in [1, 2] \)**:
\[ f'(x) = \frac{d}{dx}(2x^2 + x - 1) = 4x + 1 \]

### Step 4: Check differentiability at critical points
Now we will check the differentiability at the critical points \( x = -1, 0, 1 \):

1. **At \( x = -1 \)**:
- Left-hand derivative: \( \lim_{x \to -1^-} f'(x) = 4(-1) + 1 = -4 + 1 = -3 \)
- Right-hand derivative: \( \lim_{x \to -1^+} f'(x) = -4(-1) - 1 = 4 - 1 = 3 \)
- Since the left-hand and right-hand derivatives are not equal, \( f \) is not differentiable at \( x = -1 \).

2. **At \( x = 0 \)**:
- Left-hand derivative: \( \lim_{x \to 0^-} f'(x) = -4(0) - 1 = -1 \)
- Right-hand derivative: \( \lim_{x \to 0^+} f'(x) = 1 \)
- Since the left-hand and right-hand derivatives are not equal, \( f \) is not differentiable at \( x = 0 \).

3. **At \( x = 1 \)**:
- Left-hand derivative: \( \lim_{x \to 1^-} f'(x) = 1 \)
- Right-hand derivative: \( \lim_{x \to 1^+} f'(x) = 4(1) + 1 = 5 \)
- Since the left-hand and right-hand derivatives are not equal, \( f \) is not differentiable at \( x = 1 \).

### Conclusion
The function \( f(x) \) is not differentiable at the points \( x = -1, 0, \) and \( 1 \).