1. **Problem Statement:** We are given the function $f(x) = 1 + \frac{1}{x} + \frac{1}{x^2}$ and want to analyze its vertical and horizontal asymptotes, critical points, and inflection points. 2. **Vertical Asymptote:** Vertical asymptotes occur where the function is undefined and the limit tends to infinity. Here, $f(x)$ is undefined at $x=0$ because of division by zero in $\frac{1}{x}$ and $\frac{1}{x^2}$. So, $x=0$ is a vertical asymptote. 3. **Horizontal Asymptote:** To find horizontal asymptotes, analyze the limit of $f(x)$ as $x \to \pm \infty$: $$\lim_{x \to \infty} \left(1 + \frac{1}{x} + \frac{1}{x^2}\right) = 1 + 0 + 0 = 1$$ $$\lim_{x \to -\infty} \left(1 + \frac{1}{x} + \frac{1}{x^2}\right) = 1 + 0 + 0 = 1$$ Thus, the horizontal asymptote is $y=1$. 4. **Critical Points:** Critical points occur where the derivative $f'(x)$ is zero or undefined (except where the function is undefined). Calculate the derivative: $$f'(x) = 0 - \frac{1}{x^2} - \frac{2}{x^3} = -\frac{1}{x^2} - \frac{2}{x^3} = -\frac{x+2}{x^3}$$ Set numerator zero for critical points: $$x + 2 = 0 \implies x = -2$$ At $x=-2$, $f'(x)=0$, so this is a critical point. 5. **Local Minimum at $x=-2$:** To confirm if $x=-2$ is a local minimum, check the second derivative: $$f''(x) = \frac{2}{x^3} + \frac{6}{x^4} = \frac{2x + 6}{x^4}$$ Evaluate at $x=-2$: $$f''(-2) = \frac{2(-2) + 6}{(-2)^4} = \frac{-4 + 6}{16} = \frac{2}{16} = \frac{1}{8} > 0$$ Since $f''(-2) > 0$, $x=-2$ is a local minimum. 6. **Inflection Points:** Inflection points occur where $f''(x) = 0$ or changes sign. Set numerator of $f''(x)$ zero: $$2x + 6 = 0 \implies x = -3$$ At $x=-3$, the concavity changes from negative to positive (N shape to U shape), so $x=-3$ is an inflection point. **Final answers:** - Vertical asymptote at $x=0$ - Horizontal asymptote at $y=1$ - Local minimum at $x=-2$ - Inflection point at $x=-3$