Question

Let \( f(x+y) = f(x)f(y) \) for all \( x \) and \( y \). If \( f(0)=1, f(3)=3 \) and \( f'(0)=11 \), then \( f'(3) \) is equal to:

• A. \( 11 \)
• B. \( 22 \)
• C. \( 33 \)
• D. \( 44 \)
• E. \( 55 \)

Hint:

For exponential functions satisfying \( f(x+y)=f(x)f(y) \), the derivative is always proportional to the function itself. The constant of proportionality is always the derivative at zero: \( f'(x) = k f(x) \).

Answer 1

The Correct Option is C

Concept: The functional equation \( f(x+y) = f(x)f(y) \) is the defining property of an exponential function. Using the definition of a derivative and this property, we can establish a relationship between the derivative at any point \( x \) and the derivative at the origin.

Step 1: Express the derivative using the first principle.
\[ f'(x) = \lim_{h\to0} \frac{f(x+h) - f(x)}{h} \] Using the property \( f(x+h) = f(x)f(h) \): \[ f'(x) = \lim_{h\to0} \frac{f(x)f(h) - f(x)}{h} = f(x) \lim_{h\to0} \frac{f(h) - 1}{h} \] Since \( f(0) = 1 \), the limit term is exactly the definition of \( f'(0) \): \[ f'(x) = f(x) \cdot f'(0) \]

Step 2: Solve for \( f'(3) \).
Substitute \( x = 3 \) into the derived formula: \[ f'(3) = f(3) \cdot f'(0) \] Given \( f(3) = 3 \) and \( f'(0) = 11 \): \[ f'(3) = 3 \times 11 = 33 \]