1. **Problem Statement:** Find the intervals where the function $$y=\frac{x^2+4}{x^2-4}$$ is increasing or decreasing. 2. **Formula and Rules:** To find where a function is increasing or decreasing, we find its first derivative $$y'$$ and analyze its sign. - If $$y' > 0$$, the function is increasing. - If $$y' < 0$$, the function is decreasing. 3. **Find the first derivative:** Using the quotient rule: $$y' = \frac{(2x)(x^2-4) - (x^2+4)(2x)}{(x^2-4)^2}$$ Simplify numerator: $$= \frac{2x(x^2-4) - 2x(x^2+4)}{(x^2-4)^2} = \frac{2x x^2 - 8x - 2x x^2 - 8x}{(x^2-4)^2}$$ $$= \frac{-16x}{(x^2-4)^2}$$ 4. **Analyze the sign of $$y'$$:** - The denominator $$ (x^2-4)^2 $$ is always positive except at $$x=\pm 2$$ where the function is undefined. - The numerator is $$-16x$$. So: - $$y' > 0 \Rightarrow -16x > 0 \Rightarrow x < 0$$ - $$y' < 0 \Rightarrow -16x < 0 \Rightarrow x > 0$$ 5. **Intervals of increase and decrease:** - Increasing on $$(-\infty, -2) \cup (-2, 0)$$ (excluding points where function undefined) - Decreasing on $$(0, 2) \cup (2, \infty)$$ Note: The function is undefined at $$x=\pm 2$$, so these points split the domain. **Final answer:** - Increasing on $$(-\infty, -2) \cup (-2, 0)$$ - Decreasing on $$(0, 2) \cup (2, \infty)$$