1. The problem is to analyze the function $\rho(x) = e^{-x}$.\n\n2. This is an exponential decay function where the base of the exponent is $e$, the natural exponential constant approximately equal to 2.71828. The negative sign in the exponent indicates decay as $x$ increases.\n\n3. The general form of an exponential function is $f(x) = a e^{bx}$, where $a$ is the initial value and $b$ determines growth ($b>0$) or decay ($b<0$). Here, $a=1$ and $b=-1$.\n\n4. Important properties:\n- When $x=0$, $\rho(0) = e^0 = 1$.\n- As $x \to \infty$, $e^{-x} \to 0$.\n- As $x \to -\infty$, $e^{-x} \to \infty$.\n\n5. The function is always positive since $e^{-x} > 0$ for all real $x$.\n\n6. The derivative is $\rho'(x) = -e^{-x}$, which is always negative, confirming the function is strictly decreasing.\n\n7. Summary: $\rho(x) = e^{-x}$ starts at 1 when $x=0$, decreases towards 0 as $x$ increases, and grows without bound as $x$ decreases.