1. The problem involves analyzing the function $$f(x) = x + \frac{3}{x}$$ and its behavior as $$x \to \infty$$, finding critical points, and determining concavity. 2. First, find the derivative: $$f'(x) = 1 - \frac{3}{x^2}$$ Set $$f'(x) = 0$$ to find critical points: $$1 - \frac{3}{x^2} = 0 \implies x^2 = 3 \implies x = \pm \sqrt{3}$$ 3. Evaluate $$f(x)$$ at critical points: - At $$x = \sqrt{3}$$: $$f(\sqrt{3}) = \sqrt{3} + \frac{3}{\sqrt{3}} = \sqrt{3} + \sqrt{3} \times 3/3 = 2\sqrt{3} \approx 3.464$$ - At $$x = -\sqrt{3}$$: $$f(-\sqrt{3}) = -\sqrt{3} + \frac{3}{-\sqrt{3}} = -\sqrt{3} - \sqrt{3} = -2\sqrt{3} \approx -3.464$$ 4. Determine the nature of critical points using the second derivative: $$f''(x) = \frac{6}{x^3}$$ - For $$x > 0$$, $$f''(x) > 0$$, so the function is concave up and $$x=\sqrt{3}$$ is a local minimum. - For $$x < 0$$, $$f''(x) < 0$$, so the function is concave down and $$x=-\sqrt{3}$$ is a local maximum. 5. As $$x \to \infty$$, $$\frac{3}{x} \to 0$$, so the function approaches the asymptote $$y = x$$. Final answers: - Local minimum at $$\left(\sqrt{3}, 2\sqrt{3}\right)$$ - Local maximum at $$\left(-\sqrt{3}, -2\sqrt{3}\right)$$ - Concave up for $$x > 0$$, concave down for $$x < 0$$ - Asymptote: $$y = x$$ as $$x \to \infty$$