Problem of the Week

Problem 22 Solution

Congratulations to Cerda-Mendoza from Sawston Village College and Rian S for finding the general solution.

Any exponential function $a^x$ satisfies the first condition.
$a^{x+y}$ = $a^x \cdot a^y$ from exponent laws.

More generally the solution is $e^{kx}$ where k is a real number.

A rigorous proof can be found below. (This was not required)

$f(0+0) = (f(0))^2 = f(0)$ (since $f(0) \neq 0$ )

$\implies f(0) = 1$

$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$= \lim_{h \to 0} \frac{f(x) f(h) - f(x)}{h}$

$= f(x) \lim_{h \to 0} \frac{f(h) - 1}{h}$

$= f(x) \cdot f'(0)$

$\implies \boxed{f'(x) = f(x) f'(0)}$

$\frac{d}{dx} \left[ f(x)^a \right] = f(x)^{a-1} \cdot a \cdot f'(x)$

$= a f(x)^{a-1} f(x) f'(0)$

$= a f(x)^a f'(0)$

$\implies \left[ f(x)^a \right]' = a f(x)^a f'(0)$

$\text{Let } a = 1,$

$f(x)' = f(x) f'(0)$

This is the definition of $e^x$ (for f(x))

$\implies f(x+y) = f(x) f(y) \quad \text{generally holds when } f(x) = e^{kx}$