1. **Stating the problem:** We are given the function $$f(x) = \frac{\ln(x)}{e^x}$$ and asked to analyze it. 2. **Formula and rules:** The function is a quotient of two functions: the natural logarithm $$\ln(x)$$ and the exponential function $$e^x$$. 3. **Domain:** Since $$\ln(x)$$ is defined only for $$x > 0$$, the domain of $$f(x)$$ is $$x > 0$$. 4. **Simplification:** The function is already simplified as $$f(x) = \frac{\ln(x)}{e^x}$$. 5. **Behavior:** As $$x \to 0^+$$, $$\ln(x) \to -\infty$$ and $$e^x \to 1$$, so $$f(x) \to -\infty$$. 6. As $$x \to \infty$$, $$\ln(x)$$ grows slowly but $$e^x$$ grows very fast, so $$f(x) \to 0$$. 7. **Derivative:** To find critical points, use the quotient rule: $$ f'(x) = \frac{(\frac{1}{x}) e^x - \ln(x) e^x}{(e^x)^2} = \frac{\frac{e^x}{x} - \ln(x) e^x}{e^{2x}} = \frac{e^x (\frac{1}{x} - \ln(x))}{e^{2x}} = \frac{\frac{1}{x} - \ln(x)}{e^x} $$ 8. Set $$f'(x) = 0$$ to find critical points: $$ \frac{1}{x} - \ln(x) = 0 \implies \frac{1}{x} = \ln(x) \implies \ln(x) = \frac{1}{x} $$ 9. This transcendental equation can be solved numerically; it has a solution near $$x \approx 1.763$$. 10. **Summary:** The function $$f(x) = \frac{\ln(x)}{e^x}$$ is defined for $$x > 0$$, tends to $$-\infty$$ as $$x \to 0^+$$, tends to 0 as $$x \to \infty$$, and has a critical point near $$x \approx 1.763$$ where the derivative is zero. **Final answer:** The function $$f(x) = \frac{\ln(x)}{e^x}$$ behaves as described above with domain $$x > 0$$ and critical point near $$x \approx 1.763$$.