1. **Problem Statement:** We analyze the function $$y = x^3 + \frac{3}{x}$$ to find its critical points, concavity, and points of inflection. 2. **Find the first derivative:** $$y' = \frac{d}{dx}\left(x^3 + \frac{3}{x}\right) = 3x^2 - \frac{3}{x^2}$$ 3. **Find critical points by setting $$y' = 0$$:** $$3x^2 - \frac{3}{x^2} = 0 \implies 3x^2 = \frac{3}{x^2} \implies x^4 = 1 \implies x = \pm 1$$ 4. **Evaluate $$y$$ at critical points:** - At $$x = -1$$: $$y = (-1)^3 + \frac{3}{-1} = -1 - 3 = -4$$ - At $$x = 1$$: $$y = 1^3 + \frac{3}{1} = 1 + 3 = 4$$ 5. **Determine concavity by finding the second derivative:** $$y'' = \frac{d}{dx}\left(3x^2 - \frac{3}{x^2}\right) = 6x + \frac{6}{x^3}$$ 6. **Analyze concavity intervals:** - For $$x > 0$$, $$y'' = 6x + \frac{6}{x^3} > 0$$, so the graph is concave up on $$(0, \infty)$$. - For $$x < 0$$, $$y'' = 6x + \frac{6}{x^3} < 0$$, so the graph is concave down on $$(-\infty, 0)$$. 7. **Points of inflection:** Set $$y'' = 0$$: $$6x + \frac{6}{x^3} = 0 \implies 6x = -\frac{6}{x^3} \implies 6x^4 = -6 \implies x^4 = -1$$ No real solutions, so no points of inflection. **Final summary:** - Local maximum at $$x = -1$$ with $$y = -4$$. - Local minimum at $$x = 1$$ with $$y = 4$$. - Concave down on $$(-\infty, 0)$$. - Concave up on $$(0, \infty)$$. - No points of inflection. The function has a vertical asymptote at $$x=0$$ where it is undefined.