1. **Problem statement:** Determine which statement is true about the functions $f(x) = \left(\frac{1}{3}\right)^x$ and $g(x) = \left(\frac{3}{4}\right)^x$. 2. **Recall:** For exponential functions $a^x$ where $0 < a < 1$, the function is decreasing and $a^0 = 1$. 3. **Analyze statement A:** For $x > 0$, both $f(x)$ and $g(x)$ are powers of numbers less than 1, so both are less than 1. So statement A is true. 4. **Analyze statement B:** For $x < 0$, $f(x) = \left(\frac{1}{3}\right)^x = 3^{-x} > 1$ and $g(x) = \left(\frac{3}{4}\right)^x = \left(\frac{4}{3}\right)^{-x} > 1$, so both are greater than 1, so B is false. 5. **Analyze statement C:** For $x < 0$, $f(x) = 3^{-x}$ and $g(x) = \left(\frac{3}{4}\right)^x = \left(\frac{4}{3}\right)^{-x}$. Since $3 > \frac{4}{3}$, for negative $x$, $f(x) > g(x)$, so C is true. 6. **Analyze statement D:** For $x \geq 0$, $f(x) = \left(\frac{1}{3}\right)^x$ and $g(x) = \left(\frac{3}{4}\right)^x$. Since $\frac{1}{3} < \frac{3}{4}$ and both are less than 1, $f(x) < g(x)$ for $x > 0$, so D is false. **Final answer:** Statements A and C are true, but since the question asks which is true, the correct choice is A.