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

Answer

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

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

we will evaluate \( f(f(x)) \) by considering different cases based on the value of \( x \).

### Step 1: Case 1 - \( 0 \leq x \leq 1 \)

In this case, \( f(x) = 1 + x \).

Now we need to find \( f(f(x)) = f(1 + x) \).

- Since \( 1 + x \) will be in the range \( [1, 2] \) when \( x \) is in \( [0, 1] \), we use the first piece of the function:
\[ f(1 + x) = 1 + (1 + x) = 2 + x. \]

### Step 2: Case 2 - \( 1 < x \leq 2 \)

In this case, \( f(x) = 1 + x \).

Now we need to find \( f(f(x)) = f(1 + x) \).

- Here, \( 1 + x \) will be in the range \( (2, 3] \) when \( x \) is in \( (1, 2] \), so we use the second piece of the function:
\[ f(1 + x) = 3 - (1 + x) = 2 - x. \]

### Step 3: Case 3 - \( 2 < x \leq 3 \)

In this case, \( f(x) = 3 - x \).

Now we need to find \( f(f(x)) = f(3 - x) \).

- The value \( 3 - x \) will be in the range \( [0, 1) \) when \( x \) is in \( (2, 3] \), so we use the first piece of the function:
\[ f(3 - x) = 1 + (3 - x) = 4 - x. \]

### Final Result

Combining all the cases, we have:

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