Let f be a non-zero continuous function satisfying f(x+y)=f(x)f(y) for all , x,y `in` R . If f(2)=9 then f(3) is

Options

A. 1
B. 27
C. 9
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the value of \( f(0) \)
We start with the functional equation given:
\[ f(x+y) = f(x)f(y) \]
Let \( x = 0 \) and \( y = 0 \):
\[ f(0+0) = f(0)f(0) \implies f(0) = f(0)^2 \]
This implies:
\[ f(0)(1 - f(0)) = 0 \]
Since \( f \) is a non-zero function, we have:
\[ f(0) = 1 \]

### Step 2: Differentiate the functional equation
Next, we differentiate the equation \( f(x+y) = f(x)f(y) \) with respect to \( x \) while treating \( y \) as a constant:
\[ \frac{d}{dx}f(x+y) = \frac{d}{dx}[f(x)f(y)] \]
Using the chain rule on the left side and the product rule on the right side:
\[ f'(x+y) = f'(x)f(y) \]
Now, set \( x = 0 \):
\[ f'(0+y) = f'(0)f(y) \implies f'(y) = f'(0)f(y) \]

### Step 3: Solve the differential equation
This gives us a differential equation:
\[ \frac{f'(y)}{f(y)} = f'(0) \]
Integrating both sides with respect to \( y \):
\[ \ln|f(y)| = f'(0)y + C \]
Exponentiating both sides:
\[ f(y) = e^{f'(0)y + C} = e^C e^{f'(0)y} \]
Let \( e^C = k \), where \( k \) is a constant:
\[ f(y) = ke^{f'(0)y} \]

### Step 4: Determine the constant \( k \)
We know \( f(2) = 9 \):
\[ f(2) = ke^{2f'(0)} = 9 \]

### Step 5: Find \( f(3) \)
We need to find \( f(3) \):
\[ f(3) = ke^{3f'(0)} \]
Using the relationship we found for \( f(2) \):
\[ f(2) = ke^{2f'(0)} = 9 \implies k = 9 e^{-2f'(0)} \]
Now substituting \( k \) into the equation for \( f(3) \):
\[ f(3) = (9 e^{-2f'(0)}) e^{3f'(0)} = 9 e^{f'(0)} \]

### Step 6: Find \( e^{f'(0)} \)
From \( f(2) = 9 \):
\[ 9 = ke^{2f'(0)} \implies 9 = 9 e^{-2f'(0)} e^{2f'(0)} = 9 \]
Thus:
\[ e^{f'(0)} = 3 \]

### Step 7: Calculate \( f(3) \)
Substituting \( e^{f'(0)} \) back:
\[ f(3) = 9 \cdot 3 = 27 \]

### Final Answer
Thus, the value of \( f(3) \) is:
\[ \boxed{27} \]