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 solve the problem, we need to find \( f(f(x)) \) based on the piecewise function defined as:

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

### Step 1: Determine \( f(f(x)) \)

We will evaluate \( f(f(x)) \) by considering the two cases for \( x \) (i.e., \( x \geq 0 \) and \( x < 0 \)).

#### Case 1: \( x \geq 0 \)

1. **Calculate \( f(x) \)**:
\[ f(x) = 2 + x \]
Since \( x \geq 0 \), \( f(x) \) will also be \( \geq 2 \) (which is positive).

2. **Now calculate \( f(f(x)) \)**:
Since \( f(x) = 2 + x \geq 2 \), we use the first case of the function:
\[ f(f(x)) = f(2 + x) = 2 + (2 + x) = 4 + x \]

#### Case 2: \( x < 0 \)

1. **Calculate \( f(x) \)**:
\[ f(x) = 4 - x \]
Since \( x < 0 \), \( -x \) is positive, hence \( f(x) \) will be \( > 4 \) (which is also positive).

2. **Now calculate \( f(f(x)) \)**:
Since \( f(x) = 4 - x > 4 \), we again use the first case of the function:
\[ f(f(x)) = f(4 - x) = 2 + (4 - x) = 6 - x \]

### Final Result

Combining both cases, we have:

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