consider `f(X) =(1)/(1+|x|)+(1)/(1+|x-1|)` Let `x_(1)` and `x_(2)` be point wher f(x) attains local minmum and global maximum respectively .If `k=f(x_(1))+f(x_(2))` then 6k-9=________.

Explanation

To solve the problem step by step, we will analyze the function \( f(x) = \frac{1}{1 + |x|} + \frac{1}{1 + |x - 1|} \) and find the points where it attains local minimum and global maximum.

### Step 1: Analyze the function
The function \( f(x) \) consists of two parts: \( \frac{1}{1 + |x|} \) and \( \frac{1}{1 + |x - 1|} \). We will analyze the behavior of this function by considering the critical points and the symmetry of the function.

### Step 2: Identify critical points
The function is symmetric about \( x = 0 \) and \( x = 1 \). Therefore, we can check the behavior of the function in the intervals determined by these points, particularly between \( 0 \) and \( 1 \).

### Step 3: Find local minimum
To find the local minimum, we can evaluate \( f(x) \) at \( x = \frac{1}{2} \):
\[ f\left(\frac{1}{2}\right) = \frac{1}{1 + \frac{1}{2}} + \frac{1}{1 + \left(\frac{1}{2} - 1\right)} = \frac{1}{\frac{3}{2}} + \frac{1}{\frac{1}{2}} = \frac{2}{3} + 2 = \frac{2}{3} + \frac{6}{3} = \frac{8}{3} \]

### Step 4: Find global maximum
Next, we evaluate \( f(x) \) at the endpoints \( x = 0 \) and \( x = 1 \):
\[ f(0) = \frac{1}{1 + 0} + \frac{1}{1 + 1} = 1 + \frac{1}{2} = \frac{3}{2} \]
\[ f(1) = \frac{1}{1 + 1} + \frac{1}{1 + 0} = \frac{1}{2} + 1 = \frac{3}{2} \]
Thus, both \( f(0) \) and \( f(1) \) yield the same value, which is the global maximum.

### Step 5: Assign values to \( x_1 \) and \( x_2 \)
From the evaluations:
- Let \( x_1 = \frac{1}{2} \) (local minimum)
- Let \( x_2 = 0 \) or \( x_2 = 1 \) (global maximum)

### Step 6: Calculate \( k \)
Now, we can calculate \( k = f(x_1) + f(x_2) \):
\[ k = f\left(\frac{1}{2}\right) + f(0) = \frac{8}{3} + \frac{3}{2} \]
To add these fractions, we need a common denominator:
\[ \frac{8}{3} + \frac{3}{2} = \frac{16}{6} + \frac{9}{6} = \frac{25}{6} \]

### Step 7: Calculate \( 6k - 9 \)
Now we compute \( 6k - 9 \):
\[ 6k = 6 \times \frac{25}{6} = 25 \]
\[ 6k - 9 = 25 - 9 = 16 \]

### Final Answer
Thus, the value of \( 6k - 9 \) is \( \boxed{16} \).