1. **Problem statement:** Perform a curve discussion (Kurvendiskussion) for the function $$f(x) = \frac{4x}{x^2 - 4}$$. 2. **Domain:** The denominator cannot be zero, so solve $$x^2 - 4 = 0$$ which gives $$x = \pm 2$$. Thus, the domain is $$\mathbb{R} \setminus \{-2, 2\}$$. 3. **Intercepts:** - **y-intercept:** Set $$x=0$$, then $$f(0) = \frac{0}{-4} = 0$$. - **x-intercept:** Set $$f(x) = 0$$, so $$\frac{4x}{x^2 - 4} = 0 \Rightarrow 4x = 0 \Rightarrow x=0$$ (and $$x \neq \pm 2$$). 4. **Asymptotes:** - **Vertical asymptotes:** At points where denominator is zero, $$x = \pm 2$$. - **Horizontal asymptote:** Calculate $$\lim_{x \to \pm \infty} f(x) = \lim_{x \to \pm \infty} \frac{4x}{x^2 - 4}$$. Divide numerator and denominator by $$x^2$$: $$\lim_{x \to \pm \infty} \frac{\frac{4x}{x^2}}{\frac{x^2}{x^2} - \frac{4}{x^2}} = \lim_{x \to \pm \infty} \frac{\frac{4}{x}}{1 - \frac{4}{x^2}} = 0$$. So horizontal asymptote is $$y=0$$. 5. **First derivative:** Use quotient rule: $$f'(x) = \frac{(4)(x^2 - 4) - 4x(2x)}{(x^2 - 4)^2} = \frac{4x^2 - 16 - 8x^2}{(x^2 - 4)^2} = \frac{-4x^2 - 16}{(x^2 - 4)^2} = \frac{-4(x^2 + 4)}{(x^2 - 4)^2}$$. 6. **Critical points:** Set numerator zero: $$-4(x^2 + 4) = 0 \Rightarrow x^2 + 4 = 0$$ has no real solutions. So no critical points. 7. **Monotonicity:** Since denominator squared is always positive and numerator $$-4(x^2 + 4) < 0$$ for all real $$x$$, $$f'(x) < 0$$ everywhere in the domain. Thus, $$f(x)$$ is strictly decreasing on each interval $$(-\infty, -2)$$, $$(-2, 2)$$, and $$(2, \infty)$$. 8. **Second derivative:** Differentiate $$f'(x)$$: Let $$u = -4(x^2 + 4)$$ and $$v = (x^2 - 4)^2$$. Then $$u' = -8x$$, $$v' = 2(x^2 - 4)(2x) = 4x(x^2 - 4)$$. Using quotient rule: $$f''(x) = \frac{u'v - uv'}{v^2} = \frac{-8x (x^2 - 4)^2 - (-4(x^2 + 4)) 4x (x^2 - 4)}{(x^2 - 4)^4}$$ Simplify numerator: $$-8x (x^2 - 4)^2 + 16x (x^2 + 4)(x^2 - 4) = 8x (x^2 - 4) [ - (x^2 - 4) + 2(x^2 + 4) ]$$ Inside bracket: $$- (x^2 - 4) + 2(x^2 + 4) = -x^2 + 4 + 2x^2 + 8 = x^2 + 12$$ So numerator: $$8x (x^2 - 4)(x^2 + 12)$$ Therefore: $$f''(x) = \frac{8x (x^2 - 4)(x^2 + 12)}{(x^2 - 4)^4} = \frac{8x (x^2 + 12)}{(x^2 - 4)^3}$$. 9. **Inflection points:** Set numerator zero: $$8x (x^2 + 12) = 0 \Rightarrow x=0$$ (since $$x^2 + 12 > 0$$ always). Check if $$x=0$$ is in domain (yes). 10. **Concavity:** - For $$x < 0$$, numerator $$8x (x^2 + 12) < 0$$, denominator $$ (x^2 - 4)^3 $$ is positive for $$|x| > 2$$ and negative for $$|x| < 2$$. - Analyze intervals: - $$(-\infty, -2)$$: denominator positive, numerator negative (since $$x<0$$), so $$f''(x) < 0$$ (concave down). - $$(-2, 0)$$: denominator negative, numerator negative, so $$f''(x) > 0$$ (concave up). - $$(0, 2)$$: denominator negative, numerator positive, so $$f''(x) < 0$$ (concave down). - $$(2, \infty)$$: denominator positive, numerator positive, so $$f''(x) > 0$$ (concave up). 11. **Summary:** - Domain: $$\mathbb{R} \setminus \{-2, 2\}$$ - Intercepts: $$x=0$$, $$y=0$$ - Vertical asymptotes: $$x=\pm 2$$ - Horizontal asymptote: $$y=0$$ - Monotonically decreasing everywhere - Inflection point at $$x=0$$ - Concave down on $$(-\infty, -2) \cup (0, 2)$$ - Concave up on $$(-2, 0) \cup (2, \infty)$$