Let `x _(1) , x _(2), x _(3)` be the points where `f (x) = | 1-|x-4||, x in R` is not differentiable then `f (x_(1))+ f(x _(2)) + f (x _(3))=`

Explanation

To solve the problem, we need to find the points where the function \( f(x) = |1 - |x - 4|| \) is not differentiable and then calculate the sum \( f(x_1) + f(x_2) + f(x_3) \).

### Step-by-Step Solution:

1. **Identify the Inner Absolute Function**:
The function \( f(x) = |1 - |x - 4|| \) has an inner absolute value \( |x - 4| \). The absolute value function is not differentiable at points where its argument is zero. Therefore, we first find where \( |x - 4| = 0 \).
- Set \( x - 4 = 0 \) which gives \( x = 4 \).

2. **Determine the Points of Non-Differentiability**:
The outer absolute value \( |1 - |x - 4|| \) will also be non-differentiable where its argument is zero. We need to solve:
\[ 1 - |x - 4| = 0 \implies |x - 4| = 1 \]
This gives us two cases:
- Case 1: \( x - 4 = 1 \) → \( x = 5 \)
- Case 2: \( x - 4 = -1 \) → \( x = 3 \)

Thus, the points where \( f(x) \) is not differentiable are \( x_1 = 3 \), \( x_2 = 4 \), and \( x_3 = 5 \).

3. **Calculate \( f(x) \) at the Non-Differentiable Points**:
Now we calculate \( f(x) \) at these points:
- For \( x_1 = 3 \):
\[ f(3) = |1 - |3 - 4|| = |1 - | -1 || = |1 - 1| = |0| = 0 \]
- For \( x_2 = 4 \):
\[ f(4) = |1 - |4 - 4|| = |1 - |0|| = |1 - 0| = |1| = 1 \]
- For \( x_3 = 5 \):
\[ f(5) = |1 - |5 - 4|| = |1 - |1|| = |1 - 1| = |0| = 0 \]

4. **Sum the Values**:
Now we sum the values:
\[ f(x_1) + f(x_2) + f(x_3) = f(3) + f(4) + f(5) = 0 + 1 + 0 = 1 \]

### Final Result:
Thus, the final answer is:
\[ f(x_1) + f(x_2) + f(x_3) = 1 \]