1. **State the problem:** Determine the monotonicity of the function $$y=\frac{x^3-4x}{4}$$. 2. **Recall the formula:** To analyze monotonicity, find the first derivative $$y'$$ and determine where it is positive (increasing) or negative (decreasing). 3. **Find the derivative:** $$y=\frac{x^3-4x}{4} = \frac{1}{4}x^3 - x$$ Using the power rule: $$y' = \frac{1}{4} \cdot 3x^2 - 1 = \frac{3x^2}{4} - 1$$ 4. **Find critical points:** Set $$y' = 0$$ to find where the slope changes: $$\frac{3x^2}{4} - 1 = 0$$ Multiply both sides by 4: $$3x^2 - 4 = 0$$ $$3x^2 = 4$$ $$x^2 = \frac{4}{3}$$ $$x = \pm \frac{2}{\sqrt{3}} = \pm \frac{2\sqrt{3}}{3}$$ 5. **Test intervals:** - For $$x < -\frac{2\sqrt{3}}{3}$$, pick $$x = -2$$: $$y'(-2) = \frac{3(4)}{4} - 1 = 3 - 1 = 2 > 0$$ (increasing) - For $$-\frac{2\sqrt{3}}{3} < x < \frac{2\sqrt{3}}{3}$$, pick $$x=0$$: $$y'(0) = 0 - 1 = -1 < 0$$ (decreasing) - For $$x > \frac{2\sqrt{3}}{3}$$, pick $$x=2$$: $$y'(2) = 3 - 1 = 2 > 0$$ (increasing) 6. **Conclusion:** - The function is increasing on $$(-\infty, -\frac{2\sqrt{3}}{3})$$ and $$\left(\frac{2\sqrt{3}}{3}, \infty\right)$$. - The function is decreasing on $$\left(-\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}\right)$$. Hence, the function has local maxima and minima at $$x = \pm \frac{2\sqrt{3}}{3}$$ respectively.