1. **Problem statement:** Given the function $$y=\frac{x^2+4}{x^2-4}$$, find intervals of increase/decrease, concavity, and critical points. 2. **Find the first derivative:** Use the quotient rule: $$\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}$$ where $$u=x^2+4$$ and $$v=x^2-4$$. Calculate derivatives: $$u'=2x$$ and $$v'=2x$$. Apply quotient rule: $$y' = \frac{2x(x^2-4) - (x^2+4)(2x)}{(x^2-4)^2} = \frac{2x(x^2-4 - x^2 -4)}{(x^2-4)^2} = \frac{2x(-8)}{(x^2-4)^2} = \frac{-16x}{(x^2-4)^2}$$ 3. **Determine critical points:** Set numerator equal to zero: $$-16x=0 \implies x=0$$ Denominator zero at $$x^2-4=0 \implies x=\pm 2$$ (vertical asymptotes, not in domain). 4. **Test intervals for increasing/decreasing:** - For $$x<-2$$, pick $$x=-3$$: $$y' = \frac{-16(-3)}{(9-4)^2} = \frac{48}{25} > 0$$ increasing. - For $$-2 0$$ increasing. - For $$02$$, pick $$x=3$$: $$y' = \frac{-16(3)}{(9-4)^2} = \frac{-48}{25} < 0$$ decreasing. 5. **Find second derivative for concavity:** Differentiate $$y' = \frac{-16x}{(x^2-4)^2}$$ using quotient and chain rules. Let $$w = (x^2-4)^2$$, then $$w' = 2(x^2-4)(2x) = 4x(x^2-4)$$. Apply quotient rule: $$y'' = \frac{-16 w - (-16x) w'}{w^2} = \frac{-16 (x^2-4)^2 + 16x (4x(x^2-4))}{(x^2-4)^4}$$ Simplify numerator: $$-16 (x^2-4)^2 + 64 x^2 (x^2-4) = (x^2-4)(-16 (x^2-4) + 64 x^2)$$ Expand: $$-16 (x^2-4) + 64 x^2 = -16 x^2 + 64 + 64 x^2 = 48 x^2 + 64$$ So numerator is: $$(x^2-4)(48 x^2 + 64)$$ Therefore: $$y'' = \frac{(x^2-4)(48 x^2 + 64)}{(x^2-4)^4} = \frac{48 x^2 + 64}{(x^2-4)^3}$$ 6. **Determine concavity intervals:** - Denominator sign depends on $$x^2-4$$ cubed. - Numerator $$48 x^2 + 64 > 0$$ for all real $$x$$. Check sign of denominator: - For $$x<-2$$, $$x^2-4 > 0$$, so denominator positive cubed positive. - For $$-22$$, denominator positive cubed positive. Thus: - $$x<-2$$: $$y'' > 0$$ concave upward. - $$-22$$: $$y'' > 0$$ concave upward. 7. **Find inflection points:** Points where concavity changes and function is defined. At $$x=\pm 2$$ function undefined (vertical asymptotes), so no inflection points there. No other points where $$y''=0$$ since numerator never zero. 8. **Local maxima and minima:** - At $$x=0$$, check sign change of $$y'$$: - $$y'$$ positive to left of 0, negative to right of 0, so local maximum at $$x=0$$. Calculate $$y(0) = \frac{0+4}{0-4} = \frac{4}{-4} = -1$$. 9. **Summary:** - Increasing on $$(-\infty, -2) \cup (-2, 0)$$. - Decreasing on $$(0, 2) \cup (2, \infty)$$. - Concave upward on $$(-\infty, -2) \cup (2, \infty)$$. - Concave downward on $$(-2, 2)$$. - Local maximum at $$(0, -1)$$. - Vertical asymptotes at $$x=\pm 2$$. - No local minima or inflection points.