1. **Problem:** Let $f(x) = \left(\frac{1}{3}\right)^x$ and $g(x) = \left(\frac{3}{4}\right)^x$. Determine which of the statements A, B, C, or D is true. 2. **Recall:** For exponential functions $a^x$ where $0 < a < 1$, the function is decreasing and approaches 0 as $x \to \infty$, and approaches infinity as $x \to -\infty$. Also, $a^0 = 1$. 3. **Analyze $f(x)$:** Since $\frac{1}{3} < 1$, $f(x)$ is a decreasing exponential function. - For $x > 0$, $f(x) = \left(\frac{1}{3}\right)^x < 1$. - For $x < 0$, $f(x) = \left(\frac{1}{3}\right)^x = 3^{-x} > 1$. 4. **Analyze $g(x)$:** Since $\frac{3}{4} < 1$, $g(x)$ is also a decreasing exponential function. - For $x > 0$, $g(x) = \left(\frac{3}{4}\right)^x < 1$. - For $x < 0$, $g(x) = \left(\frac{3}{4}\right)^x = \left(\frac{4}{3}\right)^{-x} > 1$. 5. **Check statement A:** For $x > 0$, both $f(x)$ and $g(x)$ are less than 1, so both graphs lie below $y=1$. Statement A is true. 6. **Check statement B:** For $x < 0$, both $f(x)$ and $g(x)$ are greater than 1, so they lie above $y=1$. Statement B is false. 7. **Check statement C:** For $x < 0$, compare $f(x)$ and $g(x)$: $$f(x) = \left(\frac{1}{3}\right)^x = 3^{-x}, \quad g(x) = \left(\frac{3}{4}\right)^x = \left(\frac{4}{3}\right)^{-x}$$ Since $3 > \frac{4}{3}$, for negative $x$, $f(x) = 3^{-x} > \left(\frac{4}{3}\right)^{-x} = g(x)$. So $f(x)$ lies above $g(x)$ for $x < 0$. Statement C is true. 8. **Check statement D:** For $x \geq 0$, since $\frac{1}{3} < \frac{3}{4}$, for $x > 0$, $f(x) < g(x)$, so $f(x)$ lies below $g(x)$. Statement D is false. **Final conclusion:** Statements A and C are true. Since the question asks which is true, the correct choice is A (and also C is true but only one choice is expected). **Answer:** A. For $x > 0$, both graphs lie below the line $y=1$.