If `f(x)={{:(2+x",", x ge0),(4-x",", x lt 0):},` then `f (f(x))` is given by :

Options

A. `f (f(x)) ={{:(4+x",", x ge0),(6-x"," , x lt 0):}`
B. `f (f(x)) ={{:(4+x",", x ge0),(x"," , x lt 0):}`
C. `f (f(x)) ={{:(4-x",", x ge0),(x"," , x lt 0):}`
D. `f (f(x)) ={{:(4-x",", x ge0),(x+2x"," , x lt 0):}`

Answer

To find \( f(f(x)) \) for the given piecewise function

\[ f(x) = \begin{cases} 2 + x & \text{if } x \geq 0 \\ 4 - x & \text{if } x < 0 \end{cases} \]

we will evaluate \( f(f(x)) \) for both cases: when \( x \geq 0 \) and when \( x < 0 \).

### Step 1: Evaluate \( f(f(x)) \) for \( x \geq 0 \)

1. Since \( x \geq 0 \), we use the first case of the function:
\[ f(x) = 2 + x \]

2. Now we need to find \( f(f(x)) \):
\[ f(f(x)) = f(2 + x) \]

3. We need to determine if \( 2 + x \) is greater than or equal to 0. Since \( x \geq 0 \), \( 2 + x \geq 2 \) which is also greater than 0. Therefore, we use the first case again:
\[ f(2 + x) = 2 + (2 + x) = 4 + x \]

### Step 2: Evaluate \( f(f(x)) \) for \( x < 0 \)

1. Since \( x < 0 \), we use the second case of the function:
\[ f(x) = 4 - x \]

2. Now we need to find \( f(f(x)) \):
\[ f(f(x)) = f(4 - x) \]

3. We need to determine if \( 4 - x \) is greater than or equal to 0. Since \( x < 0 \), \( -x > 0 \) implies \( 4 - x > 4 \), which is greater than 0. Therefore, we again use the first case:
\[ f(4 - x) = 2 + (4 - x) = 6 - x \]

### Final Result

Combining both cases, we have:

\[ f(f(x)) = \begin{cases} 4 + x & \text{if } x \geq 0 \\ 6 - x & \text{if } x < 0 \end{cases} \]