Functional Equations
DAILY_CHALLENGE
Grade None
Question:
Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that $f(x+y) = f(x) + f(y)$ for all $x, y \in \mathbb{R}$, and $g : \mathbb{R} \to (0,\infty)$ be a function such that $g(x+y) = g(x)g(y)$ for all $x, y \in \mathbb{R}$. If $f\left(\dfrac{-3}{5}\right) = 12$ and $g\left(\dfrac{-1}{3}\right) = 2$, then the value of $\left( f\left(\dfrac{1}{4}\right) + g(-2) - 8 \right)g(0)$ is ___.
Step-by-Step Solution
Key Concept: Solutions of Cauchy's functional equations for linear and exponential models.
1) The equation $f(x+y) = f(x) + f(y)$ has the solution $f(x) = kx$.
Given $f(-3/5) = 12$:
$$k\left(-\dfrac{3}{5}\right) = 12 \implies k = -20 \implies f(x) = -20x$$
Thus, $f(1/4) = -20(1/4) = -5$.
2) The equation $g(x+y) = g(x)g(y)$ has the solution $g(x) = a^x$ (where $a > 0$).
Given $g(-1/3) = 2$:
$$a^{-1/3} = 2 \implies a = 2^{-3} = \dfrac{1}{8} \implies g(x) = \left(\dfrac{1}{8}\right)^x$$
Thus, $g(-2) = (1/8)^{-2} = 64$.
Also, $g(0) = a^0 = 1$.
3) Calculate the expression:
$$\left( f(1/4) + g(-2) - 8 \right)g(0) = (-5 + 64 - 8) \cdot 1 = 51$$
Thus, the answer is 51.
Correct Answer: 51