If `f (x)` is continous and differentiable in `[-3,9] and f'(x) in [-2,8] AA x in (-3,9).` Let N be the number of divisors of the greatest possible value of `f (9)-f (-3),` then find the sum of digits of N.

Answer

To solve the problem step by step, we will use the Mean Value Theorem and some properties of divisors.

### Step 1: Understanding the Problem
We are given that \( f(x) \) is continuous and differentiable on the interval \([-3, 9]\) and that \( f'(x) \) is bounded within the interval \([-2, 8]\) for \( x \in (-3, 9) \). We need to find the greatest possible value of \( f(9) - f(-3) \).

### Step 2: Applying the Mean Value Theorem
According to the Mean Value Theorem, there exists at least one point \( c \) in the interval \((-3, 9)\) such that:

\[ f'(c) = \frac{f(9) - f(-3)}{9 - (-3)} = \frac{f(9) - f(-3)}{12} \]

### Step 3: Finding the Maximum Value of \( f(9) - f(-3) \)
Since \( f'(x) \) is bounded by \([-2, 8]\), the maximum value of \( f'(c) \) is \( 8 \). Thus, we can set:

\[ f'(c) \leq 8 \]

Substituting this into the equation from the Mean Value Theorem gives:

\[ 8 \geq \frac{f(9) - f(-3)}{12} \]

Multiplying both sides by \( 12 \):

\[ 96 \geq f(9) - f(-3) \]

This means the greatest possible value of \( f(9) - f(-3) \) is \( 96 \).

### Step 4: Finding the Number of Divisors of \( 96 \)
Next, we need to find the number of divisors of \( 96 \). First, we factor \( 96 \):

\[ 96 = 2^5 \times 3^1 \]

### Step 5: Using the Divisor Function
The formula for the number of divisors \( N \) of a number given its prime factorization \( p_1^{a_1} \times p_2^{a_2} \times \ldots \times p_k^{a_k} \) is:

\[ N = (a_1 + 1)(a_2 + 1) \ldots (a_k + 1) \]

For \( 96 = 2^5 \times 3^1 \):

- \( a_1 = 5 \) (for \( 2^5 \))
- \( a_2 = 1 \) (for \( 3^1 \))

Thus, the number of divisors \( N \) is:

\[ N = (5 + 1)(1 + 1) = 6 \times 2 = 12 \]

### Step 6: Finding the Sum of the Digits of \( N \)
Now, we need to find the sum of the digits of \( N \):

The number \( N = 12 \) has digits \( 1 \) and \( 2 \). Therefore, the sum of the digits is:

\[ 1 + 2 = 3 \]

### Final Answer
The sum of the digits of \( N \) is \( \boxed{3} \).