1. The problem is to analyze the function $$y=\ln(x) \cdot \sin(x)$$ where $\ln(x)$ is the natural logarithm and $\sin(x)$ is the sine function. 2. Important domain rule: $\ln(x)$ is defined only for $x>0$, so the domain of $y$ is $x>0$. 3. To find critical points, we use the product rule for derivatives: $$y' = (\ln(x))' \sin(x) + \ln(x) (\sin(x))'$$ 4. Compute derivatives: $$(\ln(x))' = \frac{1}{x}$$ $$(\sin(x))' = \cos(x)$$ 5. Substitute back: $$y' = \frac{1}{x} \sin(x) + \ln(x) \cos(x)$$ 6. To find critical points, solve: $$\frac{1}{x} \sin(x) + \ln(x) \cos(x) = 0$$ 7. This equation is transcendental and generally solved numerically. 8. Intercepts: - $y$-intercept: none, since $x>0$ only. - $x$-intercepts occur when $y=0$, i.e., when $\ln(x)=0$ or $\sin(x)=0$. 9. $\ln(x)=0$ at $x=1$. 10. $\sin(x)=0$ at $x = n\pi$ for integers $n \geq 1$. 11. So $x$-intercepts are at $x=1, \pi, 2\pi, 3\pi, ...$ 12. Summary: The function is defined for $x>0$, has zeros at $x=1$ and multiples of $\pi$, and critical points satisfy $$\frac{1}{x} \sin(x) + \ln(x) \cos(x) = 0$$. Final answer: The function $y=\ln(x) \sin(x)$ is defined for $x>0$, with zeros at $x=1$ and $x=n\pi$ for $n=1,2,3,...$, and critical points found by solving $$\frac{1}{x} \sin(x) + \ln(x) \cos(x) = 0$$.