Let `f(x)={{:(1+x",", 0 le x le 2),(3-x"," ,2 lt x le 3):}`
find (fof) (x).

Answer

To find \( (f \circ f)(x) \), we need to evaluate \( f(f(x)) \). Let's break it down step by step.

### Step 1: Understand the function \( f(x) \)

The function \( f(x) \) is defined piecewise as follows:

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

### Step 2: Determine the range of \( f(x) \)

- For \( 0 \leq x \leq 2 \):
- When \( x = 0 \), \( f(0) = 1 + 0 = 1 \)
- When \( x = 2 \), \( f(2) = 1 + 2 = 3 \)

Thus, for \( 0 \leq x \leq 2 \), \( f(x) \) ranges from 1 to 3.

- For \( 2 < x \leq 3 \):
- When \( x = 2 \), \( f(2) = 3 - 2 = 1 \)
- When \( x = 3 \), \( f(3) = 3 - 3 = 0 \)

Thus, for \( 2 < x \leq 3 \), \( f(x) \) ranges from 0 to 1.

Combining these results, the overall range of \( f(x) \) is \( [0, 3] \).

### Step 3: Evaluate \( f(f(x)) \)

Now we need to evaluate \( f(f(x)) \) based on the ranges we found.

1. **Case 1**: If \( 0 \leq x \leq 2 \)
- Here, \( f(x) = 1 + x \) which lies in the interval \( [1, 3] \).
- We need to evaluate \( f(f(x)) \):
- If \( 1 \leq f(x) \leq 2 \) (i.e., \( 0 \leq x \leq 1 \)):
\[ f(f(x)) = f(1 + x) = 1 + (1 + x) = 2 + x \]
- If \( 2 < f(x) \leq 3 \) (i.e., \( 1 < x \leq 2 \)):
\[ f(f(x)) = f(1 + x) = 3 - (1 + x) = 2 - x \]

2. **Case 2**: If \( 2 < x \leq 3 \)
- Here, \( f(x) = 3 - x \) which lies in the interval \( [0, 1] \).
- We need to evaluate \( f(f(x)) \):
\[ f(f(x)) = f(3 - x) = 1 + (3 - x) = 4 - x \]

### Step 4: Combine the results

Putting it all together, we have:

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

### Final Answer

Thus, the final expression for \( (f \circ f)(x) \) is:

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