Let f be a function such that `f(x+y)=f(x)+f(y)" for all "x and y and f(x) =(2x^(2)+3x) g(x)" for all "x, " where "g(x)` is continuous and g(0) = 3. Then find f'(x)

Explanation

To find \( f'(x) \) for the function \( f \) defined by the properties given in the problem, we will follow these steps:

### Step 1: Understand the properties of the function
We are given that:
1. \( f(x+y) = f(x) + f(y) \) for all \( x \) and \( y \).
2. \( f(x) = (2x^2 + 3x) g(x) \) for all \( x \), where \( g(x) \) is continuous and \( g(0) = 3 \).

### Step 2: Use the definition of the derivative
The derivative \( f'(x) \) can be calculated using the limit definition:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 3: Substitute \( f(x+h) \) using the functional equation
From the functional equation, we can express \( f(x+h) \) as:
\[ f(x+h) = f(x) + f(h) \]
Thus, we can rewrite the derivative as:
\[ f'(x) = \lim_{h \to 0} \frac{f(x) + f(h) - f(x)}{h} = \lim_{h \to 0} \frac{f(h)}{h} \]

### Step 4: Substitute \( f(h) \) using the given expression for \( f(x) \)
Using the expression for \( f(x) \):
\[ f(h) = (2h^2 + 3h) g(h) \]
Substituting this into the limit gives:
\[ f'(x) = \lim_{h \to 0} \frac{(2h^2 + 3h) g(h)}{h} \]

### Step 5: Simplify the expression
We can simplify the limit:
\[ f'(x) = \lim_{h \to 0} \left( 2h + 3 \right) g(h) \]

### Step 6: Evaluate the limit
As \( h \to 0 \), \( g(h) \) approaches \( g(0) \), which is given as 3. Therefore:
\[ f'(x) = (2 \cdot 0 + 3) \cdot g(0) = 3 \cdot 3 = 9 \]

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