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 \( g : R \to (0, \infty) \) be a function such that \( g(x + y) = g(x)g(y) \) for all \( x, y \in \mathbb{R} \). If \( f\left(-\frac{3}{5}\right) = 12 \) and \( g\left(-\frac{1}{3}\right) = 2 \), then the value of \[ f\left(\frac{1}{4}\right) + g(-2) - 8 \cdot g(0) \] is \_\_\_\_\_.

Hint:

For functional equations, substitute known points to determine constants and simplify.

Answer 1

The functional equations imply: \[ f(x + y) = f(x) + f(y) \quad \Rightarrow \quad f(x) = kx, \] \[ g(x + y) = g(x)g(y) \quad \Rightarrow \quad g(x) = a^x. \] Given: \[ f\left(-\frac{3}{5}\right) = 12 \quad \Rightarrow \quad k\left(-\frac{3}{5}\right) = 12 \quad \Rightarrow \quad k = -20. \] Similarly: \[ g\left(-\frac{1}{3}\right) = 2 \quad \Rightarrow \quad a^{-1/3} = 2 \quad \Rightarrow \quad a = \frac{1}{8}. \] Now: \[ f\left(\frac{1}{4}\right) = -20 \cdot \frac{1}{4} = -5, \quad g(-2) = \left(\frac{1}{8}\right)^{-2} = 64, \quad g(0) = 1. \] Substituting: \[ f\left(\frac{1}{4}\right) + g(-2) - 8 \cdot g(0) = -5 + 64 - 8 \cdot 1 = 51. \]