Let `f` be a function such that `f(x+y)=f(x)+f(y)` for all `xa n dya n df(x)=(2x^2+3x)g(x)` for all`x ,` where `g(x)` is continuous and `g(0)=3.` Then find `f^(prime)(x)dot`

Explanation

To solve the problem, we need to find the derivative of the function \( f(x) \) given the conditions provided. Let's go through the steps systematically.

### Step 1: Understand the functional equation
We are given that:
\[ f(x+y) = f(x) + f(y) \quad \text{for all } x \text{ and } y. \]
This is a Cauchy functional equation, which suggests that \( f(x) \) could be a linear function.

### Step 2: Express \( f(x) \) in terms of \( g(x) \)
We are also given:
\[ f(x) = (2x^2 + 3x)g(x). \]
This means that \( f(x) \) is a product of a polynomial and a continuous function \( g(x) \).

### Step 3: Find \( f'(x) \) using the product rule
To find the derivative \( f'(x) \), we will use the product rule:
\[ f'(x) = (2x^2 + 3x)g'(x) + (g(x))(4x + 3). \]

### Step 4: Evaluate \( f'(0) \)
We need to find \( f'(x) \) at \( x = 0 \):
1. Calculate \( f(0) \):
\[ f(0) = (2(0)^2 + 3(0))g(0) = 0 \cdot g(0) = 0. \]
2. Using the functional equation, we can find \( f(h) \):
\[ f(h) = f(0 + h) = f(0) + f(h) = 0 + f(h) = f(h). \]
3. Now, substitute \( x = 0 \) in the derivative:
\[ f'(0) = (2(0)^2 + 3(0))g'(0) + g(0)(4(0) + 3). \]
This simplifies to:
\[ f'(0) = 0 \cdot g'(0) + g(0) \cdot 3 = g(0) \cdot 3. \]
4. Since \( g(0) = 3 \):
\[ f'(0) = 3 \cdot 3 = 9. \]

### Conclusion
Thus, the derivative \( f'(x) \) evaluated at \( x = 0 \) is:
\[ \boxed{9}. \]