1. **Stating the problem:** We are given the function $$f(x) = 2x + \ln(x^2 - 3)$$ and asked to analyze it by finding: a) The domain of $$f$$ b) Intervals where $$f$$ is increasing or decreasing c) Relative/local extrema d) Intervals where $$f$$ is concave up or concave down e) Points of inflection f) Asymptotes 2. **Domain of $$f$$:** The function contains $$\ln(x^2 - 3)$$, so the argument must be positive: $$x^2 - 3 > 0 \implies x^2 > 3 \implies x < -\sqrt{3} \text{ or } x > \sqrt{3}$$ Therefore, the domain is $$(-\infty, -\sqrt{3}) \cup (\sqrt{3}, \infty)$$. 3. **First derivative $$f'(x)$$ to find increasing/decreasing intervals:** $$f'(x) = \frac{d}{dx} \left(2x + \ln(x^2 - 3)\right) = 2 + \frac{1}{x^2 - 3} \cdot 2x = 2 + \frac{2x}{x^2 - 3}$$ Simplify: $$f'(x) = 2 + \frac{2x}{x^2 - 3} = \frac{2(x^2 - 3)}{x^2 - 3} + \frac{2x}{x^2 - 3} = \frac{2x^2 - 6 + 2x}{x^2 - 3} = \frac{2x^2 + 2x - 6}{x^2 - 3}$$ 4. **Critical points:** Set numerator equal to zero: $$2x^2 + 2x - 6 = 0 \implies x^2 + x - 3 = 0$$ Solve using quadratic formula: $$x = \frac{-1 \pm \sqrt{1 + 12}}{2} = \frac{-1 \pm \sqrt{13}}{2}$$ Both roots are in the domain since $$\sqrt{13} \approx 3.6$$, so: $$x_1 = \frac{-1 - 3.6}{2} = -2.3$$ (in $$(-\infty, -\sqrt{3})$$) $$x_2 = \frac{-1 + 3.6}{2} = 1.3$$ (in $$(\sqrt{3}, \infty)$$) 5. **Sign analysis of $$f'(x)$$:** Denominator $$x^2 - 3 > 0$$ in domain, so sign depends on numerator: - For $$x < -2.3$$, numerator $$> 0$$ (test $$x = -3$$: $$2(9) + 2(-3) - 6 = 18 - 6 - 6 = 6 > 0$$) - For $$-2.3 < x < -\sqrt{3}$$, numerator $$< 0$$ (test $$x = -1.8$$: $$2(3.24) + 2(-1.8) - 6 = 6.48 - 3.6 - 6 = -3.12 < 0$$) - For $$x > 1.3$$, numerator $$> 0$$ (test $$x=2$$: $$8 + 4 - 6 = 6 > 0$$) So: - Increasing on $$(-\infty, -2.3) \cup (1.3, \infty)$$ - Decreasing on $$(-2.3, -\sqrt{3}) \cup (\sqrt{3}, 1.3)$$ 6. **Relative extrema:** - At $$x = -2.3$$, $$f'$$ changes from positive to negative, so local maximum. - At $$x = 1.3$$, $$f'$$ changes from negative to positive, so local minimum. 7. **Second derivative $$f''(x)$$ for concavity:** $$f''(x) = \frac{d}{dx} f'(x) = \frac{d}{dx} \left(2 + \frac{2x}{x^2 - 3}\right) = \frac{d}{dx} \left(\frac{2x}{x^2 - 3}\right)$$ Use quotient rule: $$f''(x) = \frac{(2)(x^2 - 3) - 2x(2x)}{(x^2 - 3)^2} = \frac{2x^2 - 6 - 4x^2}{(x^2 - 3)^2} = \frac{-2x^2 - 6}{(x^2 - 3)^2}$$ 8. **Concavity intervals:** Numerator $$-2x^2 - 6 < 0$$ for all real $$x$$. Denominator $$> 0$$ in domain. So $$f''(x) < 0$$ everywhere in domain, meaning $$f$$ is concave down on $$(-\infty, -\sqrt{3}) \cup (\sqrt{3}, \infty)$$. 9. **Points of inflection:** Since $$f''(x)$$ does not change sign in domain, no points of inflection. 10. **Asymptotes:** - Vertical asymptotes at points where denominator of $$\ln$$ argument is zero or undefined: at $$x = \pm \sqrt{3}$$. - Horizontal or oblique asymptotes: As $$x \to \pm \infty$$, $$f(x) \approx 2x + \ln(x^2) = 2x + 2\ln|x|$$ which grows without bound, so no horizontal asymptotes. **Final answers:** - Domain: $$(-\infty, -\sqrt{3}) \cup (\sqrt{3}, \infty)$$ - Increasing on $$(-\infty, -2.3) \cup (1.3, \infty)$$ - Decreasing on $$(-2.3, -\sqrt{3}) \cup (\sqrt{3}, 1.3)$$ - Local max at $$x = \frac{-1 - \sqrt{13}}{2} \approx -2.3$$ - Local min at $$x = \frac{-1 + \sqrt{13}}{2} \approx 1.3$$ - Concave down everywhere in domain - No inflection points - Vertical asymptotes at $$x = \pm \sqrt{3}$$