1. **Problem Statement:** Given the function $$f(x) = \frac{1}{x+2}$$ 2. **Find the domain of definition:** The function is defined for all real numbers except where the denominator is zero. $$x + 2 \neq 0 \implies x \neq -2$$ So, the domain is $$\mathbb{R} \setminus \{-2\}$$. 3. **Find boundary limits:** - As $$x \to -2^-$$, $$f(x) = \frac{1}{x+2} \to -\infty$$. - As $$x \to -2^+$$, $$f(x) = \frac{1}{x+2} \to +\infty$$. - As $$x \to +\infty$$, $$f(x) \to 0^+$$. - As $$x \to -\infty$$, $$f(x) \to 0^-$$. 4. **Find asymptotes:** - Vertical asymptote at $$x = -2$$. - Horizontal asymptote at $$y = 0$$. 5. **Compute the first derivative:** Using the quotient rule or rewriting as $$f(x) = (x+2)^{-1}$$, $$f'(x) = -1 \cdot (x+2)^{-2} = -\frac{1}{(x+2)^2}$$. Since the denominator is squared and positive, $$f'(x) < 0$$ for all $$x \neq -2$$. 6. **Study the sign of the first derivative:** - $$f'(x) < 0$$ everywhere in the domain, so $$f(x)$$ is strictly decreasing on $$(-\infty, -2)$$ and $$(-2, +\infty)$$. 7. **Compute the second derivative:** $$f''(x) = \frac{d}{dx} \left(-\frac{1}{(x+2)^2}\right) = - \left(-2\right)(x+2)^{-3} = \frac{2}{(x+2)^3}$$. 8. **Study the sign of the second derivative:** - For $$x > -2$$, $$f''(x) > 0$$ (concave up). - For $$x < -2$$, $$f''(x) < 0$$ (concave down). 9. **Variation table:** | Interval | $$(-\infty, -2)$$ | $$(-2, +\infty)$$ | |----------------|--------------------|-------------------| | $$f'(x)$$ | Negative | Negative | | $$f(x)$$ | Decreasing | Decreasing | | $$f''(x)$$ | Negative (concave down) | Positive (concave up) | 10. **Final answer:** - Domain: $$\mathbb{R} \setminus \{-2\}$$ - Vertical asymptote: $$x = -2$$ - Horizontal asymptote: $$y = 0$$ - $$f'(x) = -\frac{1}{(x+2)^2} < 0$$, so $$f$$ is decreasing everywhere. - $$f''(x) = \frac{2}{(x+2)^3}$$ changes sign at $$x = -2$$, concave down on $$(-\infty, -2)$$ and concave up on $$(-2, +\infty)$$.