If `f (x)= {{:(1+x, 0 le x le 2),( 3x-2, 2 lt x le 3):},` then `f (f(x))` is not differentiable at:

Options

A. `x=1`
B. `x=2`
C. `x=5/2`
D. `x=3`

Answer

To find where the function \( f(f(x)) \) is not differentiable, we will analyze the function \( f(x) \) and then compute \( f(f(x)) \) step by step.

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

The function \( f(x) \) is defined as follows:
\[ f(x) = \begin{cases} 1 + x & \text{if } 0 \leq x \leq 2 \\ 3x - 2 & \text{if } 2 < x \leq 3 \end{cases} \]

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

- For \( 0 \leq x \leq 2 \):
\[ f(x) = 1 + x \quad \text{(ranges from 1 to 3)} \]
- For \( 2 < x \leq 3 \):
\[ f(x) = 3x - 2 \quad \text{(ranges from 4 to 7)} \]

Thus, the overall range of \( f(x) \) is \( [1, 3] \) for \( 0 \leq x \leq 2 \) and \( (4, 7] \) for \( 2 < x \leq 3 \).

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

#### Case 1: \( 0 \leq x \leq 2 \)

In this case, \( f(x) = 1 + x \) which lies in the interval \( [1, 3] \). Therefore, we can apply \( f \) again:
\[ f(f(x)) = f(1 + x) = \begin{cases} 1 + (1 + x) = 2 + x & \text{if } 1 + x \leq 2 \\ 3(1 + x) - 2 = 3 + 3x - 2 = 3x + 1 & \text{if } 1 + x > 2 \end{cases} \]
This means:
- If \( 0 \leq x \leq 1 \): \( f(f(x)) = 2 + x \)
- If \( 1 < x \leq 2 \): \( f(f(x)) = 3x + 1 \)

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

In this case, \( f(x) = 3x - 2 \). We need to check where this value lies:
- For \( x = 2 \): \( f(2) = 4 \)
- For \( x = 3 \): \( f(3) = 7 \)

Thus, \( f(x) \) for \( 2 < x \leq 3 \) will always be greater than 4. Therefore, we only consider:
\[ f(f(x)) = f(3x - 2) = 3(3x - 2) - 2 = 9x - 6 - 2 = 9x - 8 \]

### Step 4: Combine results

Now we have:
\[ f(f(x)) = \begin{cases} 2 + x & \text{if } 0 \leq x \leq 1 \\ 3x + 1 & \text{if } 1 < x \leq 2 \\ 9x - 8 & \text{if } 2 < x \leq 3 \end{cases} \]

### Step 5: Check for differentiability

To check where \( f(f(x)) \) is not differentiable, we need to check the points where the piecewise function changes:
- At \( x = 1 \):
- \( f(f(1)) = 2 + 1 = 3 \)
- \( f(f(1^+)) = 3(1) + 1 = 4 \)

The left-hand derivative is \( 1 \) and the right-hand derivative is \( 3 \). Since they are not equal, \( f(f(x)) \) is not differentiable at \( x = 1 \).

- At \( x = 2 \):
- \( f(f(2)) = 3(2) + 1 = 7 \)
- \( f(f(2^+)) = 9(2) - 8 = 10 \)

The left-hand derivative is \( 3 \) and the right-hand derivative is \( 9 \). Since they are not equal, \( f(f(x)) \) is not differentiable at \( x = 2 \).

### Conclusion

Thus, \( f(f(x)) \) is not differentiable at \( x = 1 \) and \( x = 2 \).

### Final Answer

The function \( f(f(x)) \) is not differentiable at \( x = 1 \) and \( x = 2 \).

---