1. **Stating the problem:** We want to analyze and graph the function $$f(x) = 2x + \ln(x^2 - 3)$$ with the given domain and properties. 2. **Domain:** The function is defined where the argument of the logarithm is positive: $$x^2 - 3 > 0 \implies x < -\sqrt{3} \text{ or } x > \sqrt{3}$$ So the domain is $$(-\infty, -\sqrt{3}) \cup (\sqrt{3}, \infty)$$. 3. **Monotonicity (Increasing/Decreasing):** The derivative is: $$f'(x) = 2 + \frac{2x}{x^2 - 3} = \frac{2(x^2 - 3) + 2x}{x^2 - 3} = \frac{2x^2 - 6 + 2x}{x^2 - 3} = \frac{2x^2 + 2x - 6}{x^2 - 3}$$ Critical points come from numerator zero: $$2x^2 + 2x - 6 = 0 \implies x^2 + x - 3 = 0$$ Using quadratic formula: $$x = \frac{-1 \pm \sqrt{1 + 12}}{2} = \frac{-1 \pm \sqrt{13}}{2}$$ Given intervals: - Increasing on $$(-\infty, \frac{1 - \sqrt{13}}{2}) \cup (\sqrt{3}, \infty)$$ - Decreasing on $$\left(\frac{1 - \sqrt{13}}{2}, -\sqrt{3}\right)$$ 4. **Local maximum:** At $$x = \frac{1 - \sqrt{13}}{2}$$, the function has a local maximum: $$f\left(\frac{1 - \sqrt{13}}{2}\right) = 2\cdot \frac{1 - \sqrt{13}}{2} + \ln\left(\left(\frac{1 - \sqrt{13}}{2}\right)^2 - 3\right) = -\sqrt{13} + \ln\left(\frac{1 + \sqrt{13}}{2}\right)$$ 5. **Concavity:** The function is concave down on the entire domain and has no points of inflection. 6. **Asymptotes:** - Vertical asymptotes at $$x = \pm \sqrt{3}$$ due to the logarithm's domain boundary. - No horizontal or oblique asymptotes. 7. **Summary:** - Domain: $$(-\infty, -\sqrt{3}) \cup (\sqrt{3}, \infty)$$ - Increasing on $$(-\infty, \frac{1 - \sqrt{13}}{2}) \cup (\sqrt{3}, \infty)$$ - Decreasing on $$\left(\frac{1 - \sqrt{13}}{2}, -\sqrt{3}\right)$$ - Local max at $$x = \frac{1 - \sqrt{13}}{2}$$ with value $$-\sqrt{13} + \ln\left(\frac{1 + \sqrt{13}}{2}\right)$$ - Concave down everywhere - Vertical asymptotes at $$x = \pm \sqrt{3}$$ - No horizontal or oblique asymptotes This fully describes the graph of $$f(x)$$.