1. **State the problem:** We have a function $$f : \mathbb{R} \to \mathbb{R}$$ defined by $$f(x) = x - \frac{x^3}{3}$$. We need to find where the function is increasing or decreasing (monotony) and produce its variation table. 2. **Calculate the derivative to study monotony:** The derivative $$f'(x)$$ indicates where $$f$$ increases or decreases. \[ f'(x) = \frac{d}{dx} \left(x - \frac{x^3}{3}\right) = 1 - x^2 \] 3. **Find critical points where $$f'(x) = 0$$:** \[ 1 - x^2 = 0 \implies x^2 = 1 \implies x = \pm 1 \] 4. **Analyze the sign of $$f'(x)$$ on intervals determined by critical points:** - For $$x < -1$$, choose $$x = -2$$: \[ f'(-2) = 1 - (-2)^2 = 1 - 4 = -3 < 0, $$ so $$f$$ is **decreasing** here. - For $$-1 < x < 1$$, choose $$x = 0$$: \[ f'(0) = 1 - 0 = 1 > 0, $$ so $$f$$ is **increasing** here. - For $$x > 1$$, choose $$x = 2$$: \[ f'(2) = 1 - 4 = -3 < 0, $$ so $$f$$ is **decreasing** here. 5. **Determine values at critical points to complete variation table:** \[ f(-1) = -1 - \frac{(-1)^3}{3} = -1 + \frac{1}{3} = -\frac{2}{3} \] \[ f(1) = 1 - \frac{1}{3} = \frac{2}{3} \] 6. **Summary in variation table form:** $$ \begin{array}{c|ccc|c} x & -\infty & & -1 & & 1 & & +\infty \\ f'(x) & & - & 0 & + & 0 & - & \\ \hline f(x) & \downarrow & & -\frac{2}{3} & \uparrow & \frac{2}{3} & \downarrow & \end{array} $$ **Final answer:** - $$f$$ is decreasing on $$(-\infty, -1)$$ - $$f$$ is increasing on $$(-1, 1)$$ - $$f$$ is decreasing on $$(1, \infty)$$