Let `f(x)={{:(1+x,0lexle1).(3-x,1ltxltoo):}`
Define f(f(x)). AIso obtain domain and range of f(f(x)).

Answer

To solve the problem, we need to find \( f(f(x)) \) for the given piecewise function \( f(x) \):

\[ f(x) = \begin{cases} 1 + x & \text{for } 0 \leq x \leq 1 \\ 3 - x & \text{for } 1 < x < \infty \end{cases} \]

### Step 1: Determine the output of \( f(x) \)

1. For \( 0 \leq x \leq 1 \):
- Here, \( f(x) = 1 + x \).
- The range of \( f(x) \) in this interval is from \( f(0) = 1 \) to \( f(1) = 2 \), so the output is \( [1, 2] \).

2. For \( 1 < x < \infty \):
- Here, \( f(x) = 3 - x \).
- As \( x \) approaches \( 1 \), \( f(x) \) approaches \( 2 \) and as \( x \) approaches infinity, \( f(x) \) approaches negative infinity. Thus, the output is \( (-\infty, 2) \).

Combining both intervals, the range of \( f(x) \) is \( (-\infty, 2) \cup [1, 2] \).

### Step 2: Find \( f(f(x)) \)

Now we need to evaluate \( f(f(x)) \) based on the output of \( f(x) \).

1. **Case 1**: If \( x \) is in \( [0, 1] \):
- Then \( f(x) = 1 + x \) which lies in \( [1, 2] \).
- Now we evaluate \( f(f(x)) \):
- Since \( f(x) \) is in \( [1, 2] \), we use the second case of \( f(x) \):
\[ f(f(x)) = f(1 + x) = 3 - (1 + x) = 2 - x \]
- Therefore, for \( 0 \leq x \leq 1 \), \( f(f(x)) = 2 - x \).

2. **Case 2**: If \( x \) is in \( (1, \infty) \):
- Then \( f(x) = 3 - x \).
- We need to determine the value of \( 3 - x \):
- For \( x > 3 \), \( 3 - x < 0 \) (which is outside the domain of \( f(x) \)).
- For \( 1 < x < 3 \), \( 3 - x \) lies in \( (0, 2) \).
- Thus, we evaluate \( f(f(x)) \):
- Since \( f(x) = 3 - x \) is in \( (0, 2) \), we use the first case of \( f(x) \):
\[ f(f(x)) = f(3 - x) = 1 + (3 - x) = 4 - x \]
- Therefore, for \( 1 < x < 3 \), \( f(f(x)) = 4 - x \).

### Step 3: Combine results

Now we can summarize the results for \( f(f(x)) \):

\[ f(f(x)) = \begin{cases} 2 - x & \text{for } 0 \leq x \leq 1 \\ 4 - x & \text{for } 1 < x < 3 \end{cases} \]

### Step 4: Determine the domain and range of \( f(f(x)) \)

- **Domain**: The domain of \( f(f(x)) \) is \( [0, 3) \) since \( f(x) \) must be valid for all \( x \) in the original function.

- **Range**:
- For \( 0 \leq x \leq 1 \), \( f(f(x)) = 2 - x \) ranges from \( 2 \) to \( 1 \).
- For \( 1 < x < 3 \), \( f(f(x)) = 4 - x \) ranges from \( 3 \) to \( 1 \).

Combining these, the overall range is \( (1, 2] \).

### Final Answer

\[ f(f(x)) = \begin{cases} 2 - x & \text{for } 0 \leq x \leq 1 \\ 4 - x & \text{for } 1 < x < 3 \end{cases} \]

- **Domain of \( f(f(x)) \)**: \( [0, 3) \)
- **Range of \( f(f(x)) \)**: \( (1, 2] \)