1. **State the problem:** Find the stationary points of the function $$y = x^3 + \frac{11}{2}x^2 - 4x + 31$$ and classify them using a sign diagram and the second derivative test. 2. **Find the first derivative:** The stationary points occur where the first derivative is zero. $$y' = \frac{d}{dx}\left(x^3 + \frac{11}{2}x^2 - 4x + 31\right) = 3x^2 + 11x - 4$$ 3. **Solve for stationary points:** Set $$y' = 0$$ $$3x^2 + 11x - 4 = 0$$ Use the quadratic formula: $$x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} = \frac{-11 \pm \sqrt{121 + 48}}{6} = \frac{-11 \pm \sqrt{169}}{6}$$ $$x = \frac{-11 \pm 13}{6}$$ So, $$x_1 = \frac{-11 + 13}{6} = \frac{2}{6} = \frac{1}{3}$$ $$x_2 = \frac{-11 - 13}{6} = \frac{-24}{6} = -4$$ 4. **Sign diagram for $$y'$$:** - Test intervals around $$x = -4$$ and $$x = \frac{1}{3}$$. - For $$x < -4$$, pick $$x = -5$$: $$y'(-5) = 3(25) + 11(-5) - 4 = 75 - 55 - 4 = 16 > 0$$ - For $$-4 < x < \frac{1}{3}$$, pick $$x = 0$$: $$y'(0) = -4 < 0$$ - For $$x > \frac{1}{3}$$, pick $$x = 1$$: $$y'(1) = 3 + 11 - 4 = 10 > 0$$ So, $$y'$$ changes from positive to negative at $$x = -4$$ (local max), and from negative to positive at $$x = \frac{1}{3}$$ (local min). 5. **Second derivative test:** Find $$y''$$: $$y'' = \frac{d}{dx} y' = 6x + 11$$ Evaluate at stationary points: - At $$x = -4$$: $$y''(-4) = 6(-4) + 11 = -24 + 11 = -13 < 0$$ (concave down, local max) - At $$x = \frac{1}{3}$$: $$y''\left(\frac{1}{3}\right) = 6 \cdot \frac{1}{3} + 11 = 2 + 11 = 13 > 0$$ (concave up, local min) 6. **Conclusion:** - Stationary points are at $$x = -4$$ (local maximum) and $$x = \frac{1}{3}$$ (local minimum). - Both sign diagram and second derivative test agree. - Preference: The second derivative test is quicker and more straightforward for classification, while the sign diagram gives a visual understanding of the function's increasing/decreasing behavior. **Final answers:** $$\boxed{\text{Local max at } x = -4, \quad \text{Local min at } x = \frac{1}{3}}$$