The functional equation is given as: \[ f(x + h) = f(x) \cdot f(h) \] Substituting \( x = 0 \) and \( h = 5 \), we get: \[ f(0 + 5) = f(0) \cdot f(5) \] Simplifying: \[ f(5) = f(0) \cdot f(5) \] Thus, we have: \[ f(5) \cdot (f(0) - 1) = 0 \] Since \( f(5) \neq 0 \), this implies: \[ f(0) = 1 \] \[\] Step 2: Finding the derivative \( f'(5) \) The derivative is defined as: \[ f'(5) = \lim_{h \to 0} \frac{f(5 + h) - f(5)}{h} \] Substituting the functional equation \( f(x + h) = f(x) \cdot f(h) \), we get: \[ f'(5) = \lim_{h \to 0} \frac{f(5) \cdot f(h) - f(5)}{h} \] Factor out \( f(5) \): \[ f'(5) = f(5) \lim_{h \to 0} \frac{f(h) - 1}{h} \] We know that \( f'(0) = 3 \), so: \[ f'(5) = f(5) \cdot f'(0) = 2 \cdot 3 = 6 \] Thus, \( f'(5) = 6 \), which matches your conclusion. \[\] Final Answer: \[ f'(5) = 6 \]