1. **State the problem:** We are given the function $$y=\frac{1}{x}$$ and asked to find the open intervals where the function is increasing, decreasing, or constant. 2. **Analyze the function:** The function $$y=\frac{1}{x}$$ is defined for all $$x \neq 0$$. It has a vertical asymptote at $$x=0$$ and a horizontal asymptote at $$y=0$$. 3. **Find the derivative:** To determine where the function is increasing or decreasing, compute the derivative: $$ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{1}{x} \right) = -\frac{1}{x^2} $$ 4. **Analyze the sign of the derivative:** Since $$x^2 > 0$$ for all $$x \neq 0$$, the derivative is always negative: $$ -\frac{1}{x^2} < 0 \quad \text{for all} \quad x \neq 0 $$ 5. **Interpretation:** Because the derivative is negative everywhere in the domain, the function is strictly decreasing on each interval where it is defined. 6. **Intervals:** The domain is $$(-\infty, 0) \cup (0, \infty)$$. On both intervals, the function is decreasing. 7. **Constant intervals:** There are no intervals where the function is constant. **Final answer:** - Increasing on: $$\varnothing$$ (no intervals) - Decreasing on: $$(-\infty, 0) \cup (0, \infty)$$ - Constant on: $$\varnothing$$ Therefore, the function is decreasing on two intervals.