Let S be the set of point where the function f (x) = |4 - |2 - x|| is non differentiable the `sum_(x in s) f (x)` =

Explanation

To find the sum of the values of the function \( f(x) = |4 - |2 - x|| \) at the points where it is non-differentiable, we will follow these steps:

### Step 1: Identify the points of non-differentiability
The function \( f(x) \) involves absolute values, which can lead to points of non-differentiability. We need to find the points where the inner absolute value \( |2 - x| \) and the outer absolute value \( |4 - |2 - x|| \) change their behavior.

1. **Find where \( |2 - x| = 0 \)**:
\[ 2 - x = 0 \implies x = 2 \]
This is one point of non-differentiability.

2. **Find where \( |4 - |2 - x|| = 0 \)**:
\[ 4 - |2 - x| = 0 \implies |2 - x| = 4 \]
This gives us two cases:
- \( 2 - x = 4 \implies x = -2 \)
- \( 2 - x = -4 \implies x = 6 \)

Thus, the points of non-differentiability are \( x = -2, 2, 6 \).

### Step 2: Evaluate \( f(x) \) at these points
Now we need to calculate \( f(x) \) at each of these points:

1. **At \( x = -2 \)**:
\[ f(-2) = |4 - |2 - (-2)|| = |4 - |2 + 2|| = |4 - |4|| = |4 - 4| = |0| = 0 \]

2. **At \( x = 2 \)**:
\[ f(2) = |4 - |2 - 2|| = |4 - |0|| = |4 - 0| = |4| = 4 \]

3. **At \( x = 6 \)**:
\[ f(6) = |4 - |2 - 6|| = |4 - |-4|| = |4 - 4| = |0| = 0 \]

### Step 3: Sum the values of \( f(x) \) at the points of non-differentiability
Now we sum the values we found:
\[ f(-2) + f(2) + f(6) = 0 + 4 + 0 = 4 \]

### Final Answer
The sum of \( f(x) \) at the points where it is non-differentiable is:
\[ \boxed{4} \]