If f is a differentiable function for all real x and `f ' (x) le 5 AA x in R.` If` f (2) = 0 and f (5) = 15` value of f (3) is _________.

Explanation

To solve the problem, we will use the Mean Value Theorem (MVT) and the properties of differentiable functions.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We know that \( f \) is a differentiable function for all real \( x \), and that \( f'(x) \leq 5 \) for all \( x \in \mathbb{R} \). We also have the values \( f(2) = 0 \) and \( f(5) = 15 \).

2. **Applying the Mean Value Theorem:**
Since \( f \) is continuous on the closed interval \([2, 5]\) and differentiable on the open interval \((2, 5)\), we can apply the Mean Value Theorem. According to MVT, there exists some \( c \in (2, 5) \) such that:
\[ f'(c) = \frac{f(5) - f(2)}{5 - 2} \]

3. **Calculating the Right-Hand Side:**
Substitute the known values into the equation:
\[ f'(c) = \frac{15 - 0}{5 - 2} = \frac{15}{3} = 5 \]

4. **Interpreting the Result:**
From the Mean Value Theorem, we have found that \( f'(c) = 5 \) for some \( c \in (2, 5) \). Since \( f'(x) \leq 5 \) for all \( x \), this means that \( f'(x) \) can be equal to 5 at some points but not exceed it.

5. **Finding the Function:**
Given that \( f'(x) \) can be at most 5, we can express \( f'(x) \) as:
\[ f'(x) = 5 + k \]
where \( k \leq 0 \) (since \( f'(x) \) cannot exceed 5).

6. **Integrating to Find \( f(x) \):**
Integrating \( f'(x) \):
\[ f(x) = 5x + C \]
where \( C \) is a constant.

7. **Using the Initial Condition to Find \( C \):**
We know \( f(2) = 0 \):
\[ 0 = 5(2) + C \implies 0 = 10 + C \implies C = -10 \]
Thus, the function is:
\[ f(x) = 5x - 10 \]

8. **Calculating \( f(3) \):**
Now we can find \( f(3) \):
\[ f(3) = 5(3) - 10 = 15 - 10 = 5 \]

### Final Answer:
The value of \( f(3) \) is \( \boxed{5} \).