If `f (x) =2+ |x| -|x-1|-|x+1|,` then `f '((1)/(2)) +f' ((3)/(2))+f' ((5)/(2))` is equal to:

Options

A. 1
B. `-1`
C. 2
D. `-2`

Answer

To solve the problem, we need to find the values of the derivative \( f'(x) \) at specific points and then sum them up. The function given is:

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

### Step 1: Determine the intervals based on the absolute values

The absolute value functions change their behavior at certain points. The critical points for \( |x| \), \( |x-1| \), and \( |x+1| \) are \( -1 \), \( 0 \), and \( 1 \). We will analyze the function in the following intervals:
- \( (-\infty, -1) \)
- \( [-1, 0) \)
- \( [0, 1) \)
- \( [1, \infty) \)

### Step 2: Evaluate \( f(x) \) in each interval

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

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

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

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

### Step 3: Calculate \( f' \) at the required points

Now we calculate \( f' \) at the points \( \frac{1}{2} \), \( \frac{3}{2} \), and \( \frac{5}{2} \):

- \( f'\left(\frac{1}{2}\right) \): Since \( \frac{1}{2} \) is in the interval \( [0, 1) \), we have \( f'\left(\frac{1}{2}\right) = 1 \).
- \( f'\left(\frac{3}{2}\right) \): Since \( \frac{3}{2} \) is in the interval \( [1, \infty) \), we have \( f'\left(\frac{3}{2}\right) = -1 \).
- \( f'\left(\frac{5}{2}\right) \): Since \( \frac{5}{2} \) is also in the interval \( [1, \infty) \), we have \( f'\left(\frac{5}{2}\right) = -1 \).

### Step 4: Sum the derivatives

Now we sum these values:
\[ f'\left(\frac{1}{2}\right) + f'\left(\frac{3}{2}\right) + f'\left(\frac{5}{2}\right) = 1 + (-1) + (-1) = 1 - 1 - 1 = -1 \]

### Final Answer

Thus, the final result is:

\[ \boxed{-1} \]