Let f(x + y) = f(x) f (y) for all x and y. Suppose that f(3) = 3 and f' (0) = 11 then f' (3) is given by:

Options

A. 22
B. 44
C. 28
D. 33

Answer

To solve the problem step by step, we can follow the reasoning provided in the video transcript:

### Step 1: Understand the functional equation
We are given the functional equation:
\[ f(x+y) = f(x) f(y) \]
for all \( x \) and \( y \). This suggests that \( f \) is an exponential function.

### Step 2: Find \( f(0) \)
Set \( x = 0 \) and \( y = 0 \):
\[ f(0 + 0) = f(0) f(0) \]
This simplifies to:
\[ f(0) = f(0)^2 \]
Let \( f(0) = a \). Then we have:
\[ a = a^2 \]
This implies \( a(a - 1) = 0 \), so \( a = 0 \) or \( a = 1 \). Since \( f(3) = 3 \) (not zero), we conclude:
\[ f(0) = 1 \]

### Step 3: Differentiate the functional equation
Differentiate both sides of the functional equation with respect to \( y \):
\[ \frac{d}{dy} f(x+y) = \frac{d}{dy} [f(x) f(y)] \]
Using the chain rule on the left side and the product rule on the right side, we get:
\[ f'(x+y) = f(x) f'(y) \]

### Step 4: Set \( y = 0 \)
Now, set \( y = 0 \):
\[ f'(x+0) = f(x) f'(0) \]
This simplifies to:
\[ f'(x) = f(x) f'(0) \]
We know \( f'(0) = 11 \), so:
\[ f'(x) = 11 f(x) \]

### Step 5: Find \( f'(3) \)
Now, we can find \( f'(3) \):
\[ f'(3) = 11 f(3) \]
We know \( f(3) = 3 \), so:
\[ f'(3) = 11 \times 3 = 33 \]

### Final Answer
Thus, the value of \( f'(3) \) is:
\[ \boxed{33} \]

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