Let f(x + y) = f(x) .f(y) ` AA x, y in R ` suppose that f(k) =3, ` k in R and f'(0) = 11` then find f'(k)

Answer

To solve the problem, we start with the functional equation given:

1. **Functional Equation**:
\[ f(x + y) = f(x) \cdot f(y) \quad \text{for all } x, y \in \mathbb{R} \]

2. **Given Values**:
\[ f(k) = 3 \quad \text{and} \quad f'(0) = 11 \]

3. **Substituting \( y = k \)**:
We substitute \( y = k \) into the functional equation:
\[ f(x + k) = f(x) \cdot f(k) \]
Since \( f(k) = 3 \), we can rewrite this as:
\[ f(x + k) = 3 \cdot f(x) \]

4. **Differentiating Both Sides**:
Now we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}[f(x + k)] = \frac{d}{dx}[3 \cdot f(x)] \]
Using the chain rule on the left side, we have:
\[ f'(x + k) = 3 \cdot f'(x) \]

5. **Substituting \( x = 0 \)**:
Next, we substitute \( x = 0 \):
\[ f'(0 + k) = 3 \cdot f'(0) \]
This simplifies to:
\[ f'(k) = 3 \cdot f'(0) \]

6. **Using the Given Value of \( f'(0) \)**:
We know that \( f'(0) = 11 \), so we substitute this value:
\[ f'(k) = 3 \cdot 11 = 33 \]

Thus, the final answer is:
\[ f'(k) = 33 \]