1. The problem is to find and graph the critical points of the function $f(x) = x^3 - x$. 2. Critical points occur where the derivative $f'(x)$ is zero or undefined. For polynomial functions, the derivative is always defined, so we only solve $f'(x) = 0$. 3. Compute the derivative: $$f'(x) = \frac{d}{dx}(x^3 - x) = 3x^2 - 1$$ 4. Set the derivative equal to zero to find critical points: $$3x^2 - 1 = 0$$ 5. Solve for $x$: $$3x^2 = 1$$ $$x^2 = \frac{1}{3}$$ $$x = \pm \frac{1}{\sqrt{3}}$$ 6. Evaluate $f(x)$ at these points: $$f\left(\frac{1}{\sqrt{3}}\right) = \left(\frac{1}{\sqrt{3}}\right)^3 - \frac{1}{\sqrt{3}} = \frac{1}{3\sqrt{3}} - \frac{1}{\sqrt{3}} = -\frac{2}{3\sqrt{3}}$$ $$f\left(-\frac{1}{\sqrt{3}}\right) = \left(-\frac{1}{\sqrt{3}}\right)^3 - \left(-\frac{1}{\sqrt{3}}\right) = -\frac{1}{3\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{2}{3\sqrt{3}}$$ 7. The critical points are: $$\left(\frac{1}{\sqrt{3}}, -\frac{2}{3\sqrt{3}}\right) \quad \text{and} \quad \left(-\frac{1}{\sqrt{3}}, \frac{2}{3\sqrt{3}}\right)$$ 8. These points can be plotted on the graph to show where the slope of the function is zero, indicating local maxima or minima. Final answer: Critical points at $x = \pm \frac{1}{\sqrt{3}}$ with corresponding $y$ values $f\left(\pm \frac{1}{\sqrt{3}}\right) = \mp \frac{2}{3\sqrt{3}}$.